On the Asymptotic Expansion of the Kashaev Invariant of the 52 Knot

On the asymptotic expansion of the Kashaev invariant of the 52 knot

Tomotada Ohtsuki

Abstract

We give a presentation of the asymptotic expansion of the Kashaev invariant of the 52 knot. As the volume conjecture states, the leading term of the expansion presents the hyperbolic volume and the Chern-Simons invariant of the complement of the 52 knot. Further, we obtain a method to compute the full Poincare asymptotics to all orders of the Kashaev invariant of the 52 knot.

1 Introduction ⟨ ⟩ ∈ C ··· In [16, 17] Kashaev defined the Kashaev invariant L N of a link L for N = 2, 3, by using the quantum dilogarithm. In [18] he conjectured that, for any hyperbolic link L, log ⟨ L ⟩ 2π · lim N = vol(S3 − L), N→∞ N where “vol” denotes the hyperbolic volume, and gave evidence for the conjecture for the figure-eight knot, the 52 knot and the 61 knot, by formal calculations. In 1999, H. ⟨ ⟩ Murakami and J. Murakami [24] proved that the Kashaev√ invariant L N of any√ link L 2π −1/N 2π −1/N is equal to the N-colored Jones polynomial JN (L; e ) of L evaluated at e , where JN (L; q) denotes the invariant obtained as the quantum invariant of links associated with the N-dimensional irreducible representation of the quantum group Uq(sl2). Further, as an extension of Kashaev’s conjecture, they conjectured that, for any knot K, √ log |J (K; e2π −1/N )| 2π · lim N = vol(S3 − K), N→∞ N where “vol” in this formula denotes the simplicial volume (normalized by multiplying by the hyperbolic volume of the regular ideal tetrahedron). This is called the volume conjecture. As a complexification of the volume conjecture, it is conjectured in [25] that, for a hyperbolic link L, √ √ log J (L; e2π −1/N ) √ 2π −1 · lim N = cs(S3 − L) + −1 vol(S3 − L) (1) N→∞ N for an appropriate choice of a branch of the logarithm, where “cs” denotes the Chern-

Simons invariant. Furthermore, it is conjectured [11] (see also [3, 12, 46]) from the view-√ 2π −1/k point of the SL(2, C) Chern-Simons theory that the asymptotic expansion of JN (K; e )

1 of a hyperbolic knot K as N, k → ∞ ﬁxing u = N/k is given by the following form, √ √ ( ∑∞ ( − ) ) 2π −1/k Nς 3/2 2π 1 i JN (K; e ) ∼ e N ω · 1 + κi · (2) N,k→∞ N u=N/k: ﬁxed i=1

for some scalars ς, ω, κi depending on K and u, though they do not discuss the Jones polynomial in the Chern-Simons theory in the case of vanishing quantum dimension, which is discussed in [38]. These conjectures look interesting in the sense that they make a bridge between quantum topology and hyperbolic geometry, and it suggests the existence of a future theory between them. An approach toward a proof of the volume conjecture has been known, which is due to Kashaev [18], Thurston [35] and Yokota [42]. We briefly review this approach, as follows. ⟨ ⟩ By definition, the Kashaev invariant K N (which we review in Section 2.1) of a knot K is given by√ a sum of fractions whose denominators are product of copies of (q)n, where 2 n q = exp(2π −1/N) and (x)n = (1 − x)(1 − x ) ··· (1 − x ); for example, the Kashaev ⟨ ⟩ invariant 52 N of the 52 knot is given by ∑ 3 ⟨ ⟩ N q 52 N = , (q) (q) − − − (q) (q) (q) 0 ≤ i,j i+j N i j 1 j j i i+j < N ( ( √ )) ∼ √N − 2πn −1/N as shown in (7). By the approximation (q)n exp 2π −1 Li2(1) Li2(e ) , we expect the following approximation, ∑ ( √ √ ) 3 N ˇ 2π −1 i/N 2π −1 j/N ⟨ 52 ⟩ ∼ N q exp √ V (e , e ) , N ? 2π −1 0 ≤ i,j i+j < N where we put ( 1 ) (1) ( 1 ) Vˇ (x, y) = Li (xy) + Li + Li (y) − Li − Li − Li (1). 2 2 xy 2 2 y 2 x 2 Further, by formally replacing the sum with an integral putting t = i/N and s = j/N, we expect that ∫ ( √ √ ) 5 N ˇ 2π −1 t 2π −1 s ⟨ 52 ⟩ ∼ N q exp √ V (e , e ) dt ds. N ?? 0 ≤ t,s 2π −1 t+s ≤ 1 Furthermore, by applying the saddle point method, we expect that the asymptotic behavior might be described by a critical value of Vˇ . Yokota [42] showed that a critical value of such a function Vˇ is given by the hyperbolic volume of the knot complement. There are problems justifying this series of arguments. The volume conjecture has been rigorously proved for some particular knots and links such as torus knots [19] (see also [5]1), the figure-eight knot (by Ekholm, see also [1]2), 1A detailed asymptotic expansion of the colored Jones polynomial for torus knots is given in [5]. 2A detailed proof of the volume conjecture for the figure-eight knot was given in [1] and the term N 3/2 in (2) was also verified there.

2 Whitehead doubles of (2, p)-torus knots [47], positive iterated torus knots [37], the 52 knot [20], and some links [9, 15, 36, 37, 41, 47]; for details see e.g. [22]. They (except for the 52 knot) have particular properties; for example, the simplicial volumes of the complements of torus knots are 0, and a critical point of Vˇ for the ﬁgure-eight knot is on the original contour of the integral. The volume conjecture for them has been proved by using such particular properties. In particular, the 52 knot is of a general case; the volume conjecture for the 52 knot has been proved by Kashaev and Yokota [20] by presenting the above mentioned sum by the residue of a certain integral. The aim of this paper is to give a presentation of the asymptotic expansion of the ⟨ ⟩ Kashaev invariant 52 N of the 52 knot rigorously (Theorem 1.1 below). Let y0 be the unique solution with positive imaginary part of (y − 1)3 = y, √ y0 = 0.3376410213... + −1 · 0.5622795120... . We put 1 x = 1 − , 0 y 0 ( ) 1 ( 1 ) ( 1 ) ( 1 ) π2 ς = √ Li (x y ) + Li + Li (y ) − Li − Li − − 2 0 0 2 x y 2 0 2 y 2 x 6 2π 1 √ 0 0 0 0 = 0.4501096100... + −1 · 0.4813049796... , √ ( ) − y0 − 1 1/2 √ ω = eπ 1/4 = 0.0901905774... − −1 · 0.6499757866... . 2y0 + 1 Then, we have ⟨ ⟩ Theorem 1.1. The asymptotic expansion of the Kashaev invariant 52 N of the 52 knot is given by the following form, √ ( d ) ∑ (2π −1) ( 1 ) ⟨ 5 ⟩ = eNς N 3/2 ω · 1 + κ · i + O , 2 N i N N d+1 i=1

for any d, where κi is some constant given by a polynomial in y0 with rational coeﬃcients; in particular, κ1 is given by ( ) 1 37 31 35 1 κ = y2 − y + + 1 = (1650y2 − 3498y + 2197) + 1 1 (2y + 1)3 4 0 6 0 8 12696 0 0 0 √ = 1.0537470859... − −1 · 0.1055728779... .

−1⟨ ⟩ −N ς −3/2 We can numerically observe that the limit of q 52 N e N tends to the above mentioned value of ω. −1⟨ ⟩ −N ς −3/2 N q 52 N e N √ 200 0.0915851738... − −1 · 0.6519891312... √ 500 0.0907489101... − −1 · 0.6507787459... √ 1000 0.0904698237... − −1 · 0.6503768725...

3 √ ⟨ ⟩ N ς 3/2 −1− − Further, we can numerically observe that the limit of ( 52 N (e N ω) 1)N/(2π 1) tends to the above mentioned value of κ1. √ ⟨ ⟩ N ς 3/2 −1 − − N ( 52 N (e N ω) 1)N/(2π 1) √ 200 1.0567234007... − −1 · 0.0885918466... √ 500 1.0549811019... − −1 · 0.0987904427... √ 1000 1.0543713307... − −1 · 0.1021833710...

As the conjecture (2) suggests, ω and κi’s of (2) are expected to be invariants of K for any hyperbolic knot√K. We have computed ω and κ1 for the 52 knot in this paper. It is conjectured that 2 −1 ω2 of a hyperbolic knot is equal to the twisted Reidemeister torsion associated with the action on sl2 of the holonomy representation of the hyperbolic structure; see Remark 1.4 below. We discuss about it for some knots in [28]. We give a proof of the theorem in Section 5 by justifying the above mentioned approach. An outline of the proof is as follows. We rewrite the sum (7) by an integral by the Poisson summation formula. When we apply the Poisson summation formula, the right-hand side of the Poisson summation formula consists of infinitely many summands, and we show that we can ignore them all except for the one at 0 in the sense that they are of sufficiently small order at N → ∞ (Proposition 4.6 and Lemma 5.8). Further, by the saddle point method (Proposition 3.5), we calculate the asymptotic expansion of the integral, and obtain the presentation of the theorem. By the method of this paper, the asymptotic behavior of the Kashaev invariant is discussed for the hyperbolic knots with up to 7 crossings in [27, 26] and for some hyperbolic knots with 8 crossings in [34]. Remark 1.2. The author has written the first version of this paper in August 2011. In February 2012, [2] was uploaded in the arXiv, in which Dimofte and Garoufalidis define a formal power series from an ideal tetrahedral decomposition of a knot complement, which is expected to be equal to the asymptotic expansion of the Kashaev invariant of the knot. √ Remark 1.3. The right-hand side of (1) is equal to 2π −1 ς, and it is called the complex volume. It is known, see e.g. [45], that the complex volume can be expressed by a critical value of the potential function. It is also known, see e.g. [10], that the complex volume can be regarded as the SL(2, C) Chern-Simons invariant. Remark 1.4. The normalization of the above mentioned Reidemeister torsion is the cohomological Reidemeister torsion associated with the meridian used in [23]. We note that the twisted Reidemeister torsion in [4] is the twisted Reidemeister torsion associated with the longitude, and it can be changed to the twisted Reidemeister torsion associated with the meridian by [29, Théorème4.1] as mentioned in [23]. The paper is organized as follows. In Section 2, we review definitions and basic properties of the notation used in this paper. In Section 3, we calculate the asymptotic expansion of Gaussian integrals by the saddle point method, and show Proposition 3.5. In Section 4, we calculate the sum corresponding to the integrals of Section 3 by the Poisson summation formula, and show Proposition 4.6. In Section 5, we give a proof of Theorem 1.1

4 by using the Poisson summation formula (Proposition 4.6) and the saddle point method (Proposition 3.5). The author would like to thank Yoshiyuki Yokota for many helpful comments on the volume conjecture and the calculation of the Kashaev invariant, and Takashi Kumagai for many helpful comments on the calculation of the saddle point method and the Pois- son summation formula. The author would also like to thank Tudor Dimofte, Stavros Garoufalidis, Rinat Kashaev, Hitoshi Murakami, Toshie Takata and Dylan Thurston for many helpful comments. He would also like to thank the referees for careful reading of the manuscript.

2 Calculation of the Kashaev invariant of the 52 knot

In this section, we review the deﬁnition of the Kashaev invariant and the calculation of the ⟨ ⟩ Kashaev invariant 52 N of the 52 knot in Section 2.1. Further, we continue to calculate ⟨ ⟩ the value of 52 N toward its asymptotic expansion in Section 2.2.

2.1 The Kashaev invariant of the 52 knot In this section, we review the deﬁnition of the Kashaev invariant of oriented knots, and ⟨ ⟩ review the calculation of the Kashaev invariant 52 N of the 52 knot, which are due to Yokota [43]. The aim of this section is to show (7) which gives the value of ⟨ 5 ⟩ . √ 2 N Let N be an integer ≥ 2. We put q = exp(2π −1/N), and put

2 n (x)n = (1 − x)(1 − x ) ··· (1 − x ) for n ≥ 0. It is known [24] that, for any n, m with n ≤ m,

(q)n(q)N−n−1 = N, (3) ∑ 1 = 1. (4) (q)m−k(q)k−n n≤k≤m Following Yokota [43],3 we review the deﬁnition of the Kashaev invariant. We put N = {0, 1, ··· ,N − 1}. For i, j, k, l ∈ N , we put

− 1 − i j 1 − i j 2 +i k 2 +j l i j N q θk l i j N q θk l Rk l = , Rk l = , (q)[i−j](q)[j−l](q)[l−k−1](q)[k−i] (q)[i−j](q)[j−l](q)[l−k−1](q)[k−i] where [m] ∈ N denotes the residue of m modulo N, and we put { 1 if [i − j] + [j − l] + [l − k − 1] + [k − i] = N − 1, θi j = k l 0 otherwise. 3We make a minor modification of the definition of weights of critical points from the definition in [43], in order to make ⟨ ⟩ K N invariant under Reidemeister moves.

5 Let K be an oriented knot. We consider a 1-tangle whose closure is isotopic to K such that its string is oriented downward at its end points. Let D be a diagram of the 1-tangle. We present D by a union of elementary tangle diagrams shown in (5). We decompose the string of D into edges by cutting it at crossings and critical points with respect to the height function of R2.A labeling is an assignment of an element of N to each edge. Here, we assign 0 to the two edges adjacent to the end points of D. For example, see (6). We deﬁne the weights of labeled elementary tangle diagrams by ( i j ) ( ) ( ) i j −1/2 W = Rk l ,W = q δk,l−1 ,W = δk,l , k l k l k l (5) ( i j ) ( i j ) ( i j ) i j 1/2 W = Rk l ,W = q δi,j+1 ,W = δi,j . k l ⟨ ⟩ Then, the Kashaev invariant K N of K is deﬁned by ∑ ∏ ∏ ⟨ ⟩ ∈ C K N = W (crossings) W (critical points) . labelings crossings critical of D points of D 4 ⟨ ⟩ Following Yokota [43], we review the calculation of the Kashaev invariant 52 N of the 52 knot, where the 52 knot is the closure of the following 1-tangle.

a k j c

i c+1 (6) b i−1 d

l e

0 e+1

We consider the above labeling. Then, it is shown by arguments in [43] that the labelings of edges adjacent to the unbounded regions vanish, i.e., a = b = c = d = e = 0. Hence, the Kashaev invariant of the 52 knot is given by ∑ ⟨ ⟩ 1/2 0 0 0 k i j 0 0 0 l 52 N = q R0 k R0 j R0 1 Ri−1 l R0 1 i,j,k,l

4Our resulting formula (7) in this section is q times the corresponding formula in [43]. This difference is because of the ⟨ ⟩ difference of the definitions of K N between ours and [43].

6 ∑ 1 −k 1 +k−j − 1 +j Nq 2 Nq 2 Nq 2 = q1/2 · · · (q)[−k](q)[k−1] (q)[−k](q)[k−j](q)[j−1] (q)[i−j](q)[j−1](q)[−i] i,j,k,l 1 −l − 1 +l × Nq 2 · Nq 2 (q)[−l](q)[l−i](q)[i−1] (q)[−l](q)[l−1] ∑ N 3q = , (q)[−k](q)[k−j](q)[j−1](q)[i−j](q)[j−1](q)[−i](q)[−l](q)[l−i](q)[i−1] i,j,k,l where we obtain the last equality from (3). Further, by (4), the above formula is rewritten,

∑ 3 ⟨ ⟩ N q 52 N = (q)i−1(q)N−i(q)j−1(q)j−1(q)i−j 1≤j≤i≤N ∑ N 3q = (q)i(q)N−i−1(q)j(q)j(q)i−j 0≤j≤i

⟨ ⟩ 2.2 Calculation of 52 N toward its asymptotic expansion ⟨ ⟩ In this section, we continue to calculate the value of the Kashaev invariant 52 N of the 52 knot toward its asymptotic expansion. ⟨ ⟩ To calculate the asymptotic expansion of 52 N , we review an integral expression of (q)n. It is known [8, 40] that ( ) ( 1 ) (2n + 1) (q)n = exp φ − φ , ( 2N 2N ) (8) ( 2n + 1) ( 1 ) (q) = exp φ 1 − − φ 1 − . n 2N 2N Here, following Faddeev [6], we deﬁne a holomorphic function φ(t) on {t ∈ C | 0 < Re t < } 1 by ∫ ∞ e(2t−1)xdx φ(t) = , −∞ 4x sinh x sinh(x/N) √ noting that this integrand has poles at nπ −1 (n ∈ Z), where, to avoid the pole at 0, we choose the following contour of the integral, { } γ = (−∞, −1 ] ∪ z ∈ C |z| = 1, Im z ≥ 0 ∪ [ 1, ∞).

We review some properties of φ(t) in Appendix A.

7 ⟨ ⟩ By using φ(t), we rewrite the presentation (7) of 52 N by (8) as ( ) ∑ (2i + 1 2j + 1) ⟨ 5 ⟩ = N 3q exp N · V˜ , , 2 N 2N 2N 0 ≤ i,j i+j < N where we put ( 1 ( 1 ) ( 1 ) V˜ (t, s) = φ t + s − + φ 1 − t − s + + φ(s) − φ(1 − s) N 2N 2N ) ( 1 ) ( 1 ) − φ(1 − t) − 3 φ + 2 φ 1 − 2N 2N 1 ( ) 1 π2 = − φ(1 − t) − 2 φ(1 − s) − √ N 2π −1 6 ( ) √ 1( 1 ) 1 1 1 − 2π −1 t + s − 2 + s2 − t − s + 2 √ 2N √ 2 2 6 5 3π −1 π −1 − log N − + . 2N 4N 4N 2 Here, we obtain the second equality by Lemmas A.2 and A.3. Hence, by putting 5 V (t, s) = V˜ (t, s) + log N 2N 1 ( ) 1 π2 = − φ(1 − t) − 2 φ(1 − s) − √ N 2π −1 6 ( ) √ √ √ 1( 1 ) 1 1 1 3π −1 π −1 − 2π −1 t + s − 2 + s2 − t − s + − + , 2 2N 2 2 6 4N 4N 2 ⟨ ⟩ the presentation of 52 N is rewritten ( ) ∑ (2i + 1 2j + 1) ⟨ 5 ⟩ = N 1/2 q exp N · V , . (9) 2 N 2N 2N 0 ≤ i,j i+j < N The range of (i/N, j/N) in this sum is given by the following domain, { } ∆ = (t, s) ∈ R2 0 ≤ t, 0 ≤ s, t + s ≤ 1 . Further, it follows from Proposition A.1 that V (t, s) converges to the following function as N → ∞,

( √ √ 2 ) ˆ 1 −2π −1 t −2π −1 s π V (t, s) = √ − Li2(e ) − 2 Li2(e ) − 2π −1 ( ) 6 (10) √ 1 1 1 −2π −1 t2 + s2 + ts − t − s + . 2 2 6 Nς We will show that the asymptotic expansion of (9) is of order e R times polynomial order in N, where ςR is the real part of ς which is given in the introduction,

ςR = Re ς = 0.4501096100... .

8 ˆ ≤ − Hence, we can ignore summands of (9) in the domain where Re V (t, s) ςR ε, since they do not contribute to the resulting expansion. Further, as we can see in Figure 1, ˆ − Re V (t, s) ςR is positive only for (t, s) in a particular subdomain of ∆. In the following lemma, we consider to restrict ∆ to a smaller domain ∆′ which includes this subdomain; this restriction will be used to verify the assumptions of the Poisson summation formula and the saddle point method later.

1.0

0.8 ∆

0.6

′ 0.4 ∆

0.2

0.2 0.4 0.6 0.8 1.0 { } ˆ ≥ Figure 1: The domain (t, s) Re V (t, s) ςR

Lemma 2.1. We put { } ∆′ = (t, s) ∈ ∆ 0.04 ≤ t ≤ 0.4, 0.05 ≤ s ≤ 0.4, t + s ≤ 0.6 . Then, the following domain { } ∈ R2 ˆ ≥ − (t, s) Re V (t, s) ςR ε (11) is included in ∆′ for some suﬃciently small ε > 0. We give a proof of the lemma in Appendix D. we can graphically observe the inclusion of the lemma in Figure 1. We put { } ∆′′ = (t, s) ∈ ∆ 0.04 ≤ t ≤ 0.9, 0.05 ≤ s ≤ 0.9 . (12)

By Lemma B.1, V (t, s) uniformly converges to Vˆ (t, s) as N → ∞ in this domain. By this uniform convergence, we can restrict ∆′′ to ∆′ later. Before this restriction, we consider ′′ ⟨ ⟩ to restrict ∆ to ∆ in the following calculation of 52 N . ⟨ ⟩ We consider the value of 52 N given in (7). The summand of the sum (7) is estimated by

N 3q N 2

= 2 , (13) (q)i+j(q)N−i−j−1(q)j(q)j(q)i (q)i(q)j n where we obtain the equality by (3). Since (q)n = (1−q )(q)n−1 from the deﬁnition of (q)n, | | ≤ ≤ N the value of 1/ (q)n is monotonically increasing with respect to n for 0 n 6 . Hence,

9 for 0 ≤ i ≤ 0.04 · N and 0 ≤ j ≤ 0.05 · N, the value of (13) is monotonically increasing with respect to i and j. Further, we can similarly show that, for 0.9 · N ≤ i < N and 0.9 · N ≤ j < N, the value of (13) is monotonically decreasing with respect to i and j. Furthermore, since V (t, s) uniformly converges to Vˆ (t, s) as N → ∞ on ∆′′ by Lemma N(ς −ε1) ′′ B.1, Re V (t, s) is bounded by e R for some ε1 > 0 on ∂∆ by Lemma 2.1. Hence, N(ς −ε2) (13) is bounded by the order e R for some ε2 > 0. Therefore, by (7),

∑ 3 N q 2 N(ς −ε ) ⟨ ⟩ R 2 52 N = + O(N e ). (q)i+j(q)N−i−j−1(q)j(q)j(q)i i,j∈Z (i/N,j/N)∈∆′′ Hence, similarly as (9), we have that ∑ ( ( )) ( ) 1/2 2i + 1 2j + 1 2 N(ς −ε ) ⟨ 5 ⟩ = N q exp N · V , + O N e R 2 2 N 2N 2N i,j∈Z (i/N,j/N)∈∆′′ ∑ ( ( ) ) ( ) Nς 1/2 2i + 1 2j + 1 2 N(ς −ε ) = e N q exp N · V , − Nς + O N e R 2 . 2N 2N i,j∈Z (i/N,j/N)∈∆′′ By Lemma 2.1, we can restrict ∆′′ to ∆′, where the error term of the sum is estimated by the order N 2e−Nε. Hence, ( ) ∑ ( ( ) ) 2i + 1 2j + 1 − ⟨ 5 ⟩ = eNς N 1/2 q exp N · V , − Nς + O(e Nε3 ) , (14) 2 N 2N 2N i,j∈Z (i/N,j/N)∈∆′ for some ε3 > 0. Further, by the Poisson summation formula, we will see that the above sum is expressed by an integral, ( ) ∫ ( ) − ⟨ ⟩ Nς 5/2 · − Nε4 52 N = e N q exp N V (t, s) Nς dt ds + O(e ) ∆′ for some ε4 > 0. We will analyse this integral in Section 5 using the saddle point method and give the proof our main Theorem 1.1 there. We will therefore devote the following two sections to recalling the saddle point method and the Poisson summation formula.

3 Calculation by the saddle point method

In this section, we calculate Gaussian integrals by the saddle point method. We calculate a Gaussian integral in Proposition 3.1, a Gaussian integral with perturbative terms in Proposition 3.2, and multi-variable cases in Propositions 3.4 and 3.5. We use Proposition 3.5 in the proof of Theorem 1.1 in Section 5. For the saddle point method, see e.g. [39].

10 Proposition 3.1. For a non-zero a ∈ C, the domain {z ∈ C | Re az2 < 0} has two connected components. We choose z0, z1 from the two components respectively. Let C be a path from z0 to z1 in C. (See Figure 2.) Then, there exists ε > 0 such that ∫ √ ( ) 2 π eN·az dz = √ √ + O e−εN , C −a · N √ √ where we choose the sign of −a such that Re z1 −a > 0.

We note that ε depends on z0, z1 and a.

√1 −a

− √1 −a z1

z0 C

Figure 2: The domain {z ∈ C | Re az2 < 0} is shaded. √ Proof. By replacing z with z/ −a, we can reduce the proof to the case where a = −1. Putting a = −1, we show that ∫ √ z1 2 π e−Nz dz = √ + O(e−εN ). z0 N ∫ √ √ ∞ −Nz2 Since −∞ e dz = π/ N, it is suﬃcient to show that ∫ ∫ z0 ∞ e−Nz2 dz = O(e−εN ), e−Nz2 dz = O(e−εN ). −∞ z1 We show the latter formula. (The former formula can be shown similarly.) The latter formula is calculated, ∫ ∫ ∫ ∞ Re z1 ∞ e−Nz2 dz = e−Nz2 dz + e−Nz2 dz, z1 z1 Re z1 and the two terms of the right-hand side are estimated, ∫ ∫ ∫ Re z1 Re z1 Re z1 2 2 2 e−Nz dz ≤ |e−Nz | · |dz| ≤ e−NRe z |dz| z1 ∫ z1 z1 Re z1 −NRe z2 −NRe z2 −εN ≤ e 1 |dz| = |Im z1| · e 1 = O(e ), z1

11 ∫ ∫ 2 ∞ ∞ −N(Re z1) − 2 − e − ≤ Nz ≤ N(Re z1)z εN 0 e dz e dz = · = O(e ) Re z1 Re z1 N Re z1 for some ε > 0. Hence, we obtain the required formula. We generalize Proposition 3.1 to the case where there are perturbative terms in the exponential of the integrand. Proposition 3.2. Let a be a non-zero complex number, and let ψ(z) and r(z) be holomorphic functions of the forms,

2 3 4 ψ(z) = az + r(z), r(z) = b3z + b4z + ··· , deﬁned in a neighborhood of 0. The domain { } z ∈ C Re ψ(z) < 0 (15) has two connected components in a neighborhood of 0. We choose z0, z1 from these two components respectively. Let C be a path from z0 to z1 in C. (See Figure 3.) Then,

∫ √ ( d ) π ∑ λ ( 1 ) eN ψ(z)dz = √ √ 1 + k + O , − · N k N d+1 C a N k=1 √ for any d, where we choose the sign of −a similarly as in Proposition 3.1, and λk’s are constants given by using coeﬃcients of the expansion of ψ(z); such presentations are obtained by formally expanding the following formula, ∞ ( ) ( ) ∑ λ ( ∂ ) u2 1 + k = exp N r exp − . (16) N k ∂u 4Na u=0 k=1

In particular, λ1 is given by 15b2 3b λ = − 3 + 4 . 1 16a3 4a2

z0 C

Figure 3: The domain {z ∈ C | Re ψ(z) < 0} is shaded.

12 Proof. We show the proposition modifying a proof of the saddle point method written in [39].5 We show the proposition, for simplicity, putting a = −1 as in the proof of Proposition 3.1. We putr ˆ(z) = r(z)/z2. Sincer ˆ(z) is analytic in a neighborhood of 0, there exists a suﬃciently small δ1 > 0 such that ∑∞ k rˆ(z) = bk+2 z i=1

for |z| < δ1. Let w be a non-negative real parameter. For each ﬁxed w, we have that ∑∞ w rˆ(z) k e = Pk(w) z k=0

for |z| < δ1, where Pk(w) is a polynomial in w of degree ≤ k. Sincer ˆ(0) = 0, there exist small δ2 > 0 and ε1 > 0 such that |rˆ(x)| ≤ ε1 for −δ2 ≤ x ≤ δ2; we can further assume that ε1 < 1 and δ2 ≤ δ1. For each ﬁxed integer m > 0, we can put ∑m w rˆ(x) k m+1 e = Pk(w) x + Rm x (17) k=0

m+1 for −δ2 ≤ x ≤ δ2 and any w ≥ 0, where Rmx is the error term, which is estimated by

dm+1 dm+1 | | ≤ w rˆ(x) | | ≤ w rˆ(x) Re Rm max Re m+1 e , Im Rm max Im m+1 e . |x| ≤ δ2 dx |x| ≤ δ2 dx Further, since dm+1 ( ) ew rˆ(x) = ew rˆ(x) · polynomial in w and diﬀerentials ofr ˆ(x) of degree ≤ m+1 , dxm+1 we have that

ε1w m+1 ε2w |Rm| ≤ e K1(w + 1) ≤ K2 e (18)

for some K1,K2 > 0 and ε2 < 1, which are independent of x and w (noting that m is bounded by using d later). Further, we replace the path C with the union of a path C1 from z0 to −δ, a path C2 from −δ to δ along the real axis, and a path C3 from δ to z1. We can assume that there exist suﬃciently small δ, ε3 > 0 such that δ ≤ δ2 and C1 and C3 are in the domain { }

z ∈ C Re ψ(z) ≤ −ε3 . (19) Then, the integral of the proposition is given by ∫ ∫ ∫ ∫ eN ψ(z)dz = eN ψ(z)dz + eN ψ(z)dz + eN ψ(z)dz. (20) C C1 C2 C3 5As for the 1-variable case, a proof of a more general statement of the saddle point method is written in [39], though the multi-variable case is not written in [39]. We review (a simpler modiﬁcation of) the proof of [39], in order to generalize it to the multi-variable case later (Proposition 3.5).

13 Since C and C are in the domain (19), we have that 1 ∫ 3 ∫ − − eN ψ(z)dz = O(e Nε3 ) and eN ψ(z)dz = O(e Nε3 ). (21) C1 C3

Hence, it is sufficient to show that the integral along C2 gives the required formula. The integral along C2 is calculated as ∫ ∫ ∫ δ δ eN ψ(z)dz = eN ψ(x)dx = e−Nx2 eNx2rˆ(x)dx C −δ −δ 2 ∫ ∫ 2∑d+1 δ δ 2 k −Nx2 2d+2 −Nx2 = Pk(Nx ) x e dx + R2d+1 x e dx, (22) − − k=0 δ δ by (17), putting w = Nx2 and m = 2d + 1. When k is odd, the summand of the first term of (22) is equal to 0, since the integrand is an odd function. When k is even, the summand of the first term of (22) is given by a sum of integrals of the following form, ∫ ∫ δ ∞ − 2 − 2 − (Nx2)l xke Nx dx = N lx2l+ke Nx dx + O(e Nε4 ) −δ −∞ ∫ ∞ 1 − 2 − = √ y2l+ke y dx + O(e Nε4 ) k/2 N N −∞ √ (2l + k − 1)!! · π − = √ + O(e Nε4 ), 2l+k/2N k/2 N for some ε4 > 0, where we obtain the first equality in a similar√ way as in the proof of Proposition 3.1, and obtain the second equality putting y = Nx. Hence, the first term of (22) is given by the following form, √ ( ∑d ) π λk − √ 1 + + O(e Nε4 ). N k N k=1 Further, by (18) putting w = Nx2, the second term of (22) is estimated by ∫ ∫ δ δ 2d+2 −Nx2 2d+2 −Nx2 R2d+1 x e dx ≤ |R2d+1| x e dx −δ −δ∫ δ ( ) 2 1 2d+2 −(1−ε2)Nx ≤ K2 x e dx = O , d+ 3 −δ N 2 where we obtain the last equality in a similar way as the above calculation. Hence, by (22), ∫ √ ( d ) π ∑ λ ( 1 ) eN ψ(z)dz = √ 1 + k + O . N k N d+1 C2 N k=1 Therefore, by (20) and (21), the integral of the proposition is given by the following form,

∫ √ ( d ) π ∑ λ ( 1 ) eN ψ(z)dz = √ 1 + k + O . N k N d+1 C N k=1

14 In particular, λ1 is concretely calculated by refining the above calculation, as follows. Since 2 rˆ(z) = b3z + b4z + ··· , we have that 1 (b ew rˆ(z) = 1 + wrˆ(z) + w2rˆ(z)2 + ··· = 1 + b wz + 3 w2 + b w)z2 + ··· . 2 3 2 4 Hence, ∫ ∫ ∞ ( ( 2 ) ) 2 b eN ψ(z)dz = e−Nx 1 + b Nx2 · x + 3 (Nx2)2 + b Nx2 x2 + ··· dx 3 2 4 C √−∞ ( ( ) ( )) π 15 2 3 1 1 = √ 1 + b + b4 + O . N 16 3 4 N N 2 This is the required formula for d = 1 when a = −1. We obtain a concrete presentation of any λk by the following formal calculation. Noting that ( ∂ ) zm = m euz , ∂u u=0 we have that ∫ ∫ ( ) 2 ∂ 2 eNaz zmdz = m eNaz +uz dz ∂u u=0 ∫ ( ) ( ∂ ) u2 = m exp Naw2 − dw ∂u 4Na u=0 √ ( ) π ( ∂ ) u2 = √ √ · m exp − , −a · N ∂u 4Na u=0 which( can be) justified∑ by a similar calculation as above, to be precise. Hence, putting ˜ m exp N r(z) = m bmz , ∫ ∫ ( ) eN ψ(z)dz = eNaz2 exp N r(z) dz ∫ ( ∑ ) Naz2 ˜ m = e bmz dz √ m ∑ ( ) ( 2 ) π ˜ ∂ m u = √ √ · bm exp − − · ∂u 4Na u=0 a N m √ ( ) ( ) π ( ∂ ) u2 = √ √ · exp N r exp − , −a · N ∂u 4Na u=0 and this gives (16). By expanding this formula formally, we obtain concrete presentations of λk’s in terms of coefficients of the expansion of ψ(z).

15 Remark 3.3. We can extend Proposition 3.2 to the case where ψ(z) depends on N in such a way that ψ(z) is of the form 1 1 1 1 ψ(z) = ψ (z) + ψ (z) + ψ (z) + ··· + ψ (z) + r (z) , (23) 0 1 N 2 N 2 m N m m N m+1 where ψi(z)’s are holomorphic functions independent of N, and we assume that ψ0(z) satisﬁes the assumption of the proposition and |rm(z)| is bounded by a constant which is independent of N. Then, eNψ(z) = ϕ(z)eNψ0(z), where we put ( 1 1 1 ) ϕ(z) = exp ψ (z) + ψ (z) + ··· + ψ (z) + r (z) 1 2 N m N m−1 m N m 1 1 1 1 = ϕ (z) + ϕ (z) + ϕ (z) + ··· + ϕ − (z) +r ˜ (z) . 0 1 N 2 N 2 m 1 N m−1 m N m

Here,r ˜m(z) is the error term such that |r˜m(z)| is bounded by a constant which is independent of N. As written in [39], in the∫ same way as in the proof of Proposition 3.2, we can Nψ0(z) show the asymptotic∫ expansion of ϕi(z)e dz by expanding ϕi(z) at z = 0. Further, Nψ0(z) we can estimate r˜m(z)e dz similarly as in the proof of Proposition 3.2, noting that only values ofr ˜m(z) in a suﬃciently small neighborhood of 0 contribute to the resulting expansion. In this way, we can justify the statement of Proposition 3.2 in the case where ψ(z) depends on N as in the form (23).

We generalize Proposition 3.1 to the case of n variables. Let A be a non-singular T n symmetric complex n×n matrix, and let z be a column vector (z1, ··· , zn) ∈ C . The domain { } z ∈ Cn Re zTA z < 0 (24) is homotopy equivalent to Sn−1. Let D be an oriented n-ball embedded in Cn such that ∂D is included in the domain (24), whose inclusion is homotopy√ equivalent. There exists a matrix P such that −A = P T P ; we note that det P = det(−A). We choose an − n-ball as a neighborhood of the origin in Rn ⊂ Cn. The matrix P 1 takes√ this n-ball to an n-ball satisfying the above assumption of D. We choose the sign of det(−A) by setting it to be det P if P takes the orientation of D to the standard orientation of Rn, and −det P otherwise. Proposition 3.4. Let A, D be as above. Then, there exists ε > 0 such that ∫ n/2 ( ) T π eN·z A zdz = √ + O e−εN , n/2 D N det(−A) √ where we put dz = dz1 ··· dzn, and we choose the sign of det(−A) as above. Proof. By changing the coordinate of z linearly, we can reduce the proof to the case where A is a diagonal matrix. Further, since d(eN·zTA zdz) = 0, we can move the domain D in Cn by Stokes’ theorem; we can also move ∂D in the domain (24) ignoring error terms of order e−Nε for some ε > 0 as in the proof of Proposition 3.1. In this way, we can move D into Rn ⊂ Cn, which means that we can reduce the proof to the product of copies of the formula of Proposition 3.1.

16 We generalize Proposition 3.4 to the case where there are perturbative terms in the exponential of the integrand. Proposition 3.5. Let A be a non-singular symmetric complex n×n matrix, and let ψ(z) and r(z) be holomorphic functions of the forms, ψ(z) = zTA z + r(z), ∑ ∑ (25) ··· ··· r(z) = r(z1, , zn) = i,j,k bijkzizjzk + i,j,k,l cijklzizjzkzl + , deﬁned in a neighborhood of 0 ∈ Cn. The restriction of the domain { } z ∈ Cn Re ψ(z) < 0 (26) to a neighborhood of 0 ∈ Cn is homotopy equivalent to Sn−1. Let D be an oriented n-ball embedded in Cn such that ∂D is included in the domain (26) whose inclusion is homotopic to a homotopy equivalence to the above Sn−1 in the domain (26). Then,

∫ ( d ) πn/2 ∑ λ ( 1 ) eN ψ(z)dz = √ 1 + i + O , n/2 − N i N d+1 D N det( A) i=1 √ for any d, where we choose the sign of det(−A) as in Proposition 3.4, and λi’s are constants given by using coeﬃcients of the expansion of ψ(z); such presentations are obtained by formally expanding the following formula, ∞ ( ) ( ) ∑ λ ( ∂ ∂ ) 1 1 + i = exp N r , ··· , exp − uT A−1u . (27) N i ∂u ∂u 4N u=0 i=1 1 n

In particular, λ1 is given by −1 ∑ 1 ∑ λ = b b a a a + c a a , 1 28 · 3 i1i2i3 i4i5i6 iσ(1)iσ(2) iσ(3)iσ(4) iσ(5)iσ(6) 32 i1i2i3i4 iτ(1)iτ(2) iτ(3)iτ(4) i1,i2,··· ,i6 i1,i2,i3,i4 σ∈S τ∈S 6 ( ) 4 −1 where Sn is the nth symmetric group, and we put aij i,j = A . Proof. Similarly as in the proof of Proposition 3.4, we can reduce the proof to the case where A = −E, where E denotes the identity matrix of size n. Then, the integral of the problem is rewritten, ∫ ( ) eN·zT z exp N r(z) dz, (28) D where we can let D be a suﬃciently small neighborhood of 0 in Rn as in the proof of Proposition 3.4. The second exponential in the integrand of (28) is calculated as ( ) 1 exp N r(z) = 1 + N r(z) + N 2r(z)2 + ··· ∑ 2 ∑ = 1 + N bijkzizjzk + N cijklzizjzkzl i,j,k i,j,k,l 1 ∑ + N 2 b b z z z z z z + ··· . 2 i1i2i3 i4i5i6 i1 i2 i3 i4 i5 i6 i1,i2,··· ,i6

17 Hence, we can calculate the asymptotic expansion of the integral (28) similarly as the calculation of λ1 in the proof of Proposition 3.2, and we can show that the asymptotic expansion of (28) is given by the following form, ∫ ( ) πn/2 λ λ N ψ(z) 1 2 ··· e dz = n/2 1 + + 2 + . D N N N We estimate the remaining part “··· ” of the above formula as in the proof of Propo- sition 3.2, as follows. We can put ∑ r(z) = zizjrîj(z) i,j for somer îj(z) which satisfies thatr îj(z) =r ˆji(z). Let wij be non-negative real parameters, and put w = (wij). For each fixed w, we have that ( ∑ ) ∑∞ ∑ ··· exp wijrîj(z) = Pi1···ik (w) zi1 zik

i,j k=0 i1,··· ,ik ≤ for sufficiently small z, where Pi1···ik (w) is a polynomial in wij’s of degree k. Further, for each fixed integer m > 0, we can put ( ∑ ) ∑m ∑ ∑ ··· ··· exp wijrîj(x) = Pi1···ik (x) xi1 xik + Rj1···jm+1 xj1 xjm+1

i,j k=0 i1,··· ,ik j1,··· ,jm+1 ··· where Rj1···jm+1 xj1 xjm+1 is the error term. Similarly as in the proof of Proposition 3.2, we can estimate it by ( ∑ ) | | ≤ Rj1···jm+1 K exp εijwij i,j for some K > 0 and εij > 0 which satisﬁes that εij = εji and E−(εij) is positive deﬁnite. Further, similarly as in the proof of Proposition 3.2, putting m = 2d + 1, we can show that the error term is of order O(1/N n/2+d+1). Hence, we can show that the integral of the problem has the asymptotic expansion of the required formula. We obtain concrete presentations of λk’s by the following formal calculation. Noting that ∂ ∂ T ··· ··· u z zi1 zim = e u=0 ∂ui1 ∂uim we have that ∫ ∫

T ∂ ∂ T T eN·z A zz ···z dz = ··· eN·z A z + u z dz i1 im ∂u ∂u u=0 ∫ i1 im( ) ∂ ∂ 1 = ··· exp N · wTA w − uTA−1u dw ∂ui ∂ui 4N u=0 1 m ( ) πn/2 ∂ ∂ 1 = √ · ··· exp − uTA−1u . n/2 N det(−A) ∂ui1 ∂uim 4N u=0

18 Hence, we obtain (27) from this formula similarly as in the proof of Proposition 3.2. The above expansion is concretely calculated as ( ) ∂ ∂ 1 ∂ ∂ 1 ( 1 ) ··· exp − uTA−1u = ··· − uTA−1u k ∂ui1 ∂ui2k 4N u=0 ∂ui1 ∂ui2k k! 4N (−1)k ∑ = a ··· a . k! 4kN k iσ(1)iσ(2) iσ(2k−1)iσ(2k) σ∈S2k

In particular, we obtain the presentation of λ1 from such calculation. Remark 3.6. Similarly as Remark 3.3, we can extend Proposition 3.5 to the case where ψ(z) depends on N in such a way that ψ(z) is of the form 1 1 1 1 ψ(z) = ψ (z) + ψ (z) + ψ (z) + ··· + ψ (z) + r (z) , (29) 0 1 N 2 N 2 m N m m N m+1

where ψi(z)’s are holomorphic functions independent of N, and we assume that ψ0(z) satisﬁes the assumption of the proposition and |rm(z)| is bounded by a constant which is independent of N.

4 Calculation by the Poisson summation formula

In this section, we calculate the sums corresponding to the integrals of the propositions in the previous section by the Poisson summation formula. Corresponding to Propositions 3.1, 3.2, 3.4 and 3.5 in the previous section, we show Propositions 4.1, 4.2, 4.5 and 4.6 in this section. We use Proposition 4.6 in the proof of Theorem 1.1 in Section 5. We remark that the Poisson summation formula has also been used in the study of large level asymptotics of quantum invariants of Seifert 3-manifolds [13, 14, 30, 31, 32]. Recall (see e.g. [33]) that the Poisson summation formula states that ∑ ∑ f(m) = fˆ(m) (30) m ∈ Zn m ∈ Zn for a continuous integrable function f on Rn which satisﬁes that ( )− − ( )− − |f(z)| ≤ C 1 + |z| n δ, |fˆ(z)| ≤ C 1 + |z| n δ (31)

for some C, δ > 0, where fˆ is the Fourier transform of f deﬁned by ∫ √ fˆ(w) = f(z) e−2π −1 wT zdz. Rn Proposition 4.1. Let a, c be complex numbers satisfying that π 1 Re a < 0, Im c < − Re . 2 a We put { } k k Λ = + c ∈ C k ∈ Z, b ≤ ≤ b , N 0 N 1

19 { }

C = t + c ∈ C t ∈ R, b0 ≤ t ≤ b1 for some b0, b1 ∈ R and ε0 > 0 satisfying that Re (b0 + c) < 0 < Re (b1 + c) and Re a(bi + 2 c) < −ε0 (i = 0, 1); see Figure 4. Then, there exists ε > 0 such that ∫ ∑ ( ) 1 2 2 eN·az = eN·az dz + O e−εN . N z ∈ Λ C

We note that ε depends on a, c and ε0.

b1 +c b0 +c Λ

Figure 4: The domain {z ∈ C | Re az2 < 0} is shaded.

Proof. By Proposition 3.1, the required formula is rewritten, √ √ ∑ ( ) 2 π · N eN·a(k/N+c) = √ + O e−εN . −a k ∈ Z b0 ≤ k/N ≤ b1 Since ∑ ( ) ∑ ( ) eN·a(k/N+c)2 = O e−εN , eN·a(k/N+c)2 = O e−εN , k ∈ Z k ∈ Z k/N < b0 b1 < k/N for some ε > 0, it is suﬃcient to show that √ √ ∑ ( ) 2 π · N eN·a(k/N+c) = √ + O e−εN . (32) −a k ∈ Z

In order to apply the Poisson summation formula (30), we put f(z) = eN·a(z/N+c)2 . Then, its Fourier transform is ∫ ∞ √ fˆ(ζ) = eN·a(z/N+c)2−2π −1 ζzdz ∫−∞ ∞ √ √ = e(a/N)(z+cN−π −1ζN/a)2 dz · eNπ2ζ2/a + 2π −1 cNζ √−∞ √ √ π · N 2 2 = √ eNπ ζ /a + 2π −1 cNζ , −a

20 where we obtain the last equality by Proposition 3.1. These f(z) and fˆ(ζ) satisfy the assumption (31) of the Poisson summation formula. Hence, by the Poisson summation formula (30), √ √ ∑ ∑ √ 2 π · N 2 2 eN·a(k/N+c) = √ eNπ m /a + 2π −1 cNm. −a k ∈ Z m ∈ Z The summand at m = 0 gives the right-hand side of (32). When m ≠ 0, the power of the summand is √ (π 2 −1 c) N · πm2 + , a m and its real part is negative by the assumption of the proposition. Hence, the summands at m ≠ 0 are of order e−εN for some ε > 0. Therefore, we obtain (32). We generalize Proposition 4.1 to the case where there are perturbative terms in the exponential of the summand. Proposition 4.2. We put { } k k Λ = + c ∈ C k ∈ Z, b0 ≤ ≤ b1 , {N N }

C = t + c ∈ C t ∈ R, b0 ≤ t ≤ b1

for some b0, b1 ∈ R and c ∈ C. Let a be a complex number whose real part is negative, let δ0 be a real positive number, and let ψ(z) be a holomorphic function of the form,

2 3 4 ψ(z) = az + b3z + b4z + ··· ,

deﬁned in a neighborhood of 0 including the δ0-neighborhood of C. The domain (15) has two connected components in a neighborhood of 0. We assume that b0 + c and b1 + c are in these two components respectively, and Re ψ(bi + c) < −ε0 (i = 0, 1) for some ε0 > 0. (See Figure 5.) Further, we assume that

{b0 + c and√ b1 + c are in the same connected( component√ of ) } (33) w + δ −1 ∈ C w ∈ C, δ ∈ [0, δ0], Re ψ(w + δ −1) − 2πδ < 0 , and

{they are√ in the same connected component( of √ ) } (34) w − δ −1 ∈ C w ∈ C, δ ∈ [0, δ0], Re ψ(w − δ −1) − 2πδ < 0 . Then, ∫ 1 ∑ eN ψ(z) = eN ψ(z)dz + O(e−Nε) N z∈Λ C for some ε > 0. ′ We note that ε depends on ψ(z), c, δ0 and ε0; in particular, ε directly depends on ε2 of (39) in the proof of the proposition.

21 b1 +c b0 +c Λ

Figure 5: The domain {z ∈ C | Re ψ(z) < 0} is shaded.

Proof. The sum of the left-hand side of the required formula is rewritten, ( ) ∑ ( k ) exp N · ψ + c . (35) N k ∈ Z b0 ≤ k/N ≤ b1 In order to apply the Poisson summation formula (30), we put ( ) ( z ) ( z ) f(z) = g + c exp N · ψ + c , N N where g is a diﬀerentiable function on R + c satisfying that { 1 if w ∈ C, g(w) = 0 if w∈ / N(C), 0 ≤ g(w) ≤ 1 if w ∈ N(C) − C. {Here, N (C) is a neighborhood} of C in R+c such that N(C)−C is included in the domain

z ∈ C Re ψ(z) < −ε0/2 . Then, the Fourier transform of f is given by ∫ ( ) ( z ) ( z ) √ fˆ(ζ) = g + c exp N · ψ + c e−2π −1 ζzdz R∫ N N ( √ ) = N g(w) eN ψ(w) −2π −1 ζ(w−c) dw, R+c where we put w = z/N + c. Further, ∫ ( ) 1 ( d ) √ ζ2fˆ(ζ) = − g(w) eN ψ(w) 2e−2π −1 Nζ(w−c) dw 4π2N dw ∫R+c ( ) 1 ( d ) √ − 2 N ψ(w) −2π −1 Nζ(w−c) = 2 g(w) e e dw 4π N ∫R+c dw 1 √ − N ψ(w) −2π −1 Nζ(w−c) = 2 h(w) e e dw, 4π N R+c where we put ( ) h(w) = g′′(w) + 2 g′(w) ψ′(w) + g(w) ψ′′(w) + ψ′(w)2 .

22 Since the above integral is bounded independently of ζ, fˆ(ζ) satisﬁes the assumption (31) of the Poisson summation formula. Further, f(z) also satisﬁes (31). Therefore, by the Poisson summation formula (30), ∑ (35) = fˆ(m). m ∈ Z When m ≠ 0, we have that ∫ ( ) 1 1 √ fˆ(m) = − · h(w) eN ψ(w)−2π −1 m(w−c) dw 4π2N m2 ∫R+c ( ) 1 1 ( ) √ − · ′′ ′ 2 N ψ(w)−2π −1 m(w−c) = 2 2 ψ (w) + ψ (w) e dw (36) 4π N m ∫C ( √ ) − 1 · 1 N ψ(w)−2π −1 m(w−c) 2 2 h(w) e dw, (37) 4π N m N(C)−C since h(w) = ψ′′(w) + ψ′(w)2 for w ∈ C and h(w) = 0 for w ∈ (R + c) − N(C). Further, since Re ψ(w) < −ε /2 for w ∈ N(C) − C, 0 ∑ − (37) = O(e Nε1 ) m=0̸ ′ for some ε1 > 0. Furthermore, when m > 0, by pushing the contour C to a contour C in the domain of (34), we can show that ∫ ( ) ( ) √ ψ′′(w) + ψ′(w)2 eN ψ(w)−2π −1 m(w−c) dw C ∫ ( ) ( ) √ ′′ ′ − − − − = ψ (w) + ψ (w)2 eN ψ(w) 2π 1 m(w c) dw = O(e Nε2 ) (38) C′ ′ for some ε2 > 0, which we can choose independently of m, since there exists ε2 > 0 such that ( √ ) − − − − ′ Re ψ(w) 2π 1 (w c) < ε2 (39) ′ for any w ∈ C . Hence, ∑ − (36) = O(e Nε2 ). m>0 When m < 0, by pushing the contour C into the domain of (33), we similarly obtain ∑ − (36) = O(e Nε3 ) m<0 for some ε3 > 0. Therefore, ∑ − fˆ(m) = O(e Nε4 ) m=0̸ for some ε > 0. Hence, 4 ∫ − − (35) = fˆ(0) + O(e Nε4 ) = N eN ψ(w)dw + O(e Nε5 ) C for some ε5 > 0, and this implies the required formula.

23 Remark 4.3. The assumptions (33) and (34) of Proposition 4.2 can be replaced with the condition that there exist positive integers m0, m1 and positive real numbers δ0, δ1 such that for any integer m satisfying that −m0 < m < 0 or 0 < m < m1, ∫ ( √ ) eN ψ(w)−2π −1 m(w−c) dw = O(e−Nε) C for some ε > 0, and

{b0 + c and√ b1 + c are in the same connected( component√ of ) }

w + δ −1 ∈ C w ∈ C, δ ∈ [0, δ1], Re ψ(w + δ −1) − 2πδm1 < 0 , and

{they are√ in the same connected component( of √ ) }

w − δ −1 ∈ C w ∈ C, δ ∈ [0, δ0], Re ψ(w − δ −1) − 2πδm0 < 0 . Proposition 4.2 for this assumption can be proved by modifying the above proof of Propo- sition 4.2. Remark 4.4. Proposition 4.2 can naturally be extended to the case where the holomorphic function ψ(z) depends on N, if ψ(z) uniformly converges to ψ0(z) as N → ∞, and ′′ ′ 2 ψ0(z) satisﬁes the assumption of the proposition, and |ψ (z)+ψ (z) | is bounded by a constant which is independent of N. In this case, (39) holds for suﬃciently large N, and hence, (38) holds. Therefore, Proposition 4.2 also holds in this case, where we note that we can choose ε independently of N.

We generalize Proposition 4.1 to the case of n variables. Proposition 4.5. For c ∈ Cn and an oriented n-ball D′ in Rn, we put { } 1 1 Λ = k + c ∈ Cn k ∈ Zn, k ∈ D′ , {N N} D = z + c ∈ Cn z ∈ D′ ⊂ Rn . Let A be a non-singular symmetric complex n×n matrix such that Re (A−1) is negative deﬁnite. We assume that ∂D is included in the domain (24). Further, we assume that π (Im c)T m < − mT · Re (A−1) · m 2 for any m ∈ Zn −{0}. Then, ∑ ∫ 1 T T eN·z A z = eN·z A zdz + O(e−Nε), N n z ∈ Λ D for some ε > 0. Proof. By Proposition 3.4, the required formula is rewritten, ∑ n/2 n/2 ( ) T π N eN·z A z = √ + O e−εN . (40) − z ∈ Λ det( A)

24 Further, in a similar way as in the proof of Proposition 4.1, putting f(z) = eN·(z/N+c)T A (z/N+c), we can show that n/2 n/2 √ π N 2 T −1 T fˆ(ζ) = √ eNπ ζ A ζ + 2π −1Nc ζ det(−A) by using Proposition 3.4. Hence, we obtain

∑ n/2 n/2 ∑ √ T π N 2 T −1 T eN·(k/N+c) A (k/N+c) = √ eNπ m A m + 2π −1Nc m − k ∈ Zn det( A) m ∈ Zn by the Poisson summation formula (30). By the assumption of the proposition, the real part of √ Nπ2 mT A−1m + 2π −1NcT m is negative for any m ∈ Zn −{0}. Therefore, only the summand at m = 0 survives, and we obtain (40) from it. We generalize Proposition 4.5 to the case where there are perturbative terms in the n exponential of the summand. Let e1, e2, ··· , en be the standard basis vectors of C ,

T T T e1 = (1, 0, ··· , 0) , e2 = (0, 1, 0, ··· , 0) , ··· , en = (0, ··· , 0, 1) . Proposition 4.6. For c ∈ Cn and an oriented n-ball D′ in Rn, we put { } 1 1 Λ = k + c ∈ Cn k ∈ Zn, k ∈ D′ , {N N} D = z + c ∈ Cn z ∈ D′ ⊂ Rn .

Let A be a non-singular symmetric complex n×n matrix, and let ψ(z) be a holomorphic function of the form (25) deﬁned in{ a neighborhood of 0 ∈ C}n including D. We assume n that ∂D is included in the domain z ∈ C | Re ψ(z) < −ε0 for some ε0 > 0. Further, ′ ··· we assume that there exist δi, δi > 0 (i = 1, , n) such that

{∂D is null-homotopic√ in ( √ ) } n (41) w + δ −1 ei ∈ C w ∈ D, δ ∈ [0, δi], Re ψ(w + δ −1 ei) − 2πδ < 0 , and ∂D is null-homotopic in { √ ( √ ) } (42) − − ∈ Cn ∈ ∈ ′ − − − w δ 1 ei w D, δ [0, δi], Re ψ(w δ 1 ei) 2πδ < 0 , for i = 1, ··· , n, assuming that ψ is holomorphic in these domains. Then, ∫ 1 ∑ eN ψ(z) = eN ψ(z)dz + O(e−Nε), N n z ∈ Λ D for some ε > 0. ′ We note that ε depends on ψ(z), c, δi, δi and ε0.

25 Proof. The sum of the left-hand side of the required formula is rewritten, ( ) ∑ ( 1 ) exp N · ψ k + c . (43) N k ∈ Zn k/N ∈ D′ In order to apply the Poisson summation formula (30), we put ( ) ( 1 ) ( 1 ) f(z) = g z + c exp N · ψ z + c , N N where g is a diﬀerentiable function on Rn + c satisfying that { 1 if w ∈ D, g(w) = 0 if w ∈/ N(D), 0 ≤ g(w) ≤ 1 if w ∈ N(D) − D.

Here, N{(D) is a neighborhood of D} in Rn + c such that N(D) − D is included in the n domain z ∈ C | Re ψ(z) < −ε0/2 . Then, the Fourier transform of f is given by ∫ ( ) ( ( )) √ 1 1 T fˆ(ζ) = g z + c exp N · ψ z + c e−2π −1 ζ zdz Rn ∫ N N ( √ ) T = N n g(w) eN ψ(w)−2π −1 ζ (w−c) dw, Rn+c where we put w = z/N + c. Further, for an integer l > n/2, ( ∑n ) | |2l ˆ 2 l ˆ ζ f(ζ) = ζi f(ζ) i=1 ∫ ( − ) (( ∑n 2 ) √ ) n 1 l N ψ(w) ∂ l −2π −1 NζT (w−c) = N 2 g(w) e 2 e dw 4π N Rn+c ∂wi ∫ i=1 ( − ) (( ∑n 2 ) ) √ n 1 l ∂ l N ψ(w) −2π −1 NζT (w−c) = N 2 2 g(w) e e dw 4π N Rn ∂w ∫ +c i=1 i ( − ) √ n 1 l N ψ(w) −2π −1 NζT (w−c) = N 2 h(w) e e dw, 4π N Rn+c where h(w) is some polynomial in derivatives of g(w) and ψ(w). Since the above integral is bounded independently of ζ, fˆ(ζ) satisﬁes the assumption (31) of the Poisson summation formula. Further, f(z) also satisﬁes (31). Therefore, by the Poisson summation formula (30), ∑ (43) = fˆ(m). m ∈ Zn When m ≠ 0, we have that ∫ ( ) ( ) √ −1 1 T fˆ(m) = N n l · h(w) eN ψ(w)−2π −1 m (w−c) dw 2 | |2l 4π N m Rn+c

26 ∫ ( ) ( ) √ −1 1 T = N n l · Ψ(w) eN ψ(w)−2π −1 m (w−c) dw (44) 2 | |2l 4π N m D∫ ( ) ( ) √ −1 1 T + N n l · h(w) eN ψ(w)−2π −1 m (w−c) dw, (45) 2 | |2l 4π N m N(D)−D where Ψ(w) is some polynomial in (at most the 2lth) derivatives of ψ(w). Further, since Re ψ(w) < 0 for w ∈ N(D) − D, ∑ − (45) = O(e Nε1 ) m≠ 0 for some ε1 > 0. Furthermore, when m1 > 0, pushing the contour D into the domain of (42), we obtain ∑ − (44) = O(e Nε2 ) m ∈ Zn m1>0 for some ε2 > 0, similarly as in the proof of Proposition 4.2. Similarly we obtain ∑ − (44) = O(e Nε3 ) m ∈ Zn m1<0 for some ε3 > 0, from the assumption (41). Hence, ∑ ∑ − (44) = (44) + O(e Nε4 ) m≠ 0 m≠ 0 m1=0 for some ε4 > 0. By repeating this argument for m2, ··· , mn, we obtain ∑ − (44) = O(e Nε5 ) m≠ 0 for some ε5 > 0. Therefore, ∫ − − (43) = fˆ(0) + O(e Nε6 ) = N n eN ψ(w)dw + O(e Nε6 ) D for some ε6 > 0, and this implies the required formula. Remark 4.7. The assumptions (41) and (42) of Proposition 4.6 can be modiﬁed similarly as in Remark 4.3. Remark 4.8. Similarly as in Remark 4.4, Proposition 4.6 can naturally be extended to the case where the holomorphic function ψ(z) depends on N, if ψ(z) uniformly converges to ψ0(z) as N → ∞, and ψ0(z) satisﬁes the assumption of the proposition, and |Ψ(z)| is bounded by a constant which is independent of N. Similarly as in Remark 4.4, we can also choose ε independently of N in this case.

27 5 Proof of Theorem 1.1

In this section, we give a proof of Theorem 1.1. In Section 5.1, we give a proof of the theorem by using lemmas proved in Sections 5.2, 5.3 and 5.4. In Section 5.2, we show lemmas which calculate the asymptotic expansion. In Section 5.3, we show a lemma which veriﬁes the assumption of the saddle point method. In Section 5.4, we show a lemma which veriﬁes the assumption of the Poisson summation formula.

5.1 Proof of Theorem 1.1 In this section, we give a proof of Theorem 1.1, which presents the asymptotic expansion ⟨ ⟩ of the Kashaev invariant 52 N of the 52 knot. To obtain the asymptotic expansion, as ⟨ ⟩ mentioned in Section 2.2, we rewrite the sum (14) of 52 N by using an integral by the Poisson summation formula (Proposition 4.6) and calculate the asymptotic expansion of the integral by the saddle point method (Proposition 3.5). ⟨ ⟩ Proof of Theorem 1.1. We recall that 52 N is presented by the sum (14). By Proposition 4.6 (Poisson summation formula) (see also Remark 4.8 and Appendix B), this sum is expressed by an integral, ( ) ∫ ( ) − ⟨ ⟩ Nς 5/2 · − Nε4 52 N = e N q exp N V (t, s) Nς dt ds + O(e ) (46) ∆′ for some ε4 > 0, noting that we verify the assumptions of Proposition 4.6 in Lemma 5.8 below. In order to apply the saddle point method (Proposition 3.5, see also Remark 3.6 and Appendix B) to (46), we consider a critical point of Vˆ (t, s). From the deﬁnition of Vˆ (t, s), a critical point is a solution of the following equations, ∂ √ √ ( 1) Vˆ (t, s) = − log(1 − e−2π −1 t) − 2π −1 t + s − = 0, ∂t 2 (47) ∂ √ √ ( ) Vˆ (t, s) = −2 log(1 − e−2π −1 s) − 2π −1 t + 2s − 1 = 0. ∂s √ √ Hence, putting x = e2π −1 t and y = e2π −1 s, ( 1 ) ( 1) 1 − xy = −1, 1 − 2xy2 = 1. x y Therefore, 1 x = 1 − , (y − 1)3 = y. y

Let (x0, y0) be the solution of these equations mentioned in the√ introduction.√ We consider ′ 2π −1 t0 2π −1 s0 the critical point (t0, s0) such that (Re t0, Re s0) ∈ ∆ and (e , e ) = (x0, y0). It is numerically given by √ √ t0 = 0.22404487... − −1 · 0.04475430... , s0 = 0.16393269... + −1 · 0.06713145... .

28 Let (tc, sc) be the critical point of V (t, s) which goes to (t0, s0) as N → ∞. We consider to apply Proposition 3.5 to V (t − tc, s − sc). We consider the expansion 1 1 V (t, s) = V (t , s ) + V · (t − t )2 + V · (t − t )(s − s ) + V · (s − s )2 + ··· (48) c c 2 tt c ts c c 2 ss c at the critical point, where we put ∂2V ∂2V ∂2V V = (t , s ),V = (t , s ),V = (t , s ). tt ∂t2 c c ts ∂t ∂s c c ss ∂s2 c c As we show in Section 5.3, we can make a concrete homotopy which moves ∆′ to a new domain containing the above critical point in such a way that it satisﬁes the assumption of the saddle point method.6 Hence, we obtain the following expansion by applying Proposition 3.5 (see also Remark 3.6 and Appendix B) to (46),

( ) ( ∑d ) 2π ( )− ⟨ 5 ⟩ = N 5/2 q exp NV (t , s ) · V V − V 2 1/2 1 + λ ℏi + O(ℏd+1) , (49) 2 N c c N tt ss ts i i=1 where λi’s are given in the proposition, noting that we verify the assumptions of Propo- − sition 3.5 in Lemma( 5.7 below.) Here, we also note that eNς × O(e Nε4 ) in (46) is in- × ℏd+1 cluded( in exp NV) (tc, sc) O( ) in (49); see Lemma 5.2 below for the behavior of exp NV (tc, sc) . We calculate the concrete form of (49), for simplicity, when d = 1. By Lemmas 5.2, 5.3 and 5.4 below, (49) is rewritten, ( ) √ ( ) 2π ⟨ 5 ⟩ = N 5/2q × eNς e−π −1/4 1 − y 1/2 1 + (C + C )ℏ + O(ℏ2) × 2 N 0 1 2 N ( ) ( ) ( ) 1 −1/2 2 2 × √ − 2y0 − 1 1 + C3ℏ + O(ℏ ) × 1 + (C4 + C5)ℏ + O(ℏ ) 2π −1 √ ( ) ( ( ) ) Nς 3/2 π −1/4 2y0 + 1 −1/2 2 = e N e 1 + ℏ 1 + C1 + C2 + C3 + C4 + C5 + O(ℏ ) , y0 − 1 where the constants C1, ··· ,C5 are given in the lemmas. In particular, κ1 of the theorem is given by

κ1 = 1 + C1 + C2 + C3 + C4 + C5 1 1 1 ( ) = (−16y2 + 33y + 9) + (2y2 − 5y ) + 8y2 − 17y + 9 184 0 0 24 0 0 46 0 0 1 1 + (2y2 + 15y ) − (197y2 + 127y + 227) + 1 92 0 0 3174 0 0 1 = (1650y2 − 3498y + 2197) + 1. 12696 0 0 Hence, we obtain the required formula for d = 1.

6 ˆ ≤ − That is, the new domain is in the area Re V (t, s) ςR ε except for a neighborhood of the above critical point. In√ order to obtain this√ new domain, it is suﬃcient to push the shaded part in Figure 1 into the imaginary direction by (− −1 · 0.04475430..., −1 · 0.06713145...). For details, see Section 5.3.

29 For general d, each λi of the expansion (49) is given by a linear sum of higher coefficients of the expansion (48) by Proposition 3.5 (see also Remark 3.6 and Appendix B). In the same way as the above case of d = 1, such coefficients are given by some polynomials in y0 with rational coefficients. Hence, each κi of the theorem can be given by some polynomial in y0 with rational coefficients.

5.2 Calculation of the asymptotic expansion In this section, we show some lemmas used in the proof of Theorem 1.1. ˆ Let (t0, s0) and (tc, sc) be√ the critical points√ of V (t, s) and V (√t, s) given in the√ previous 2π −1 ·t0 2π −1 ·s0 2π −1 ·tc 2π −1 ·sc section. We put x0 = e , y0 = e and xc = e , yc = e . In order to show Lemma 5.2 below, we calculate (xc, yc) in terms of (x0, y0), as follows. We show a proof of the following lemma in Appendix C. 2 2 Lemma 5.1. We can put xc = x0 + x1ℏ + O(ℏ ) and yc = y0 + y1ℏ + O(ℏ ) for some x1, y1 ∈ C in a sufficiently small neighborhood of (x0, y0)(i.e., for sufficiently large N). Here, O(ℏ2) means that the absolute value of the error term is bounded by Cℏ2 with a constant C which is independent of N. √ √ 2π −1 ·t 2π −1 ·s We calculate x1 and y1, as follows. We put x = e and y = e . From the definition of V (t, s) and Proposition A.1, we have that ∂ ( 1 ) √ ( 1 1 ) V (t, s) = − log 1 − − 2π −1 t + s − − + O(ℏ2), ∂t x 2 2N (50) ∂ ( 1) √ ( 1 ) V (t, s) = −2 log 1 − − 2π −1 t + 2s − 1 − + O(ℏ2). ∂s y 2N

Then, since (tc, sc) is a critical point of V (t, s), we have that  ( )  1 −1/2 2  1 − (−q xcyc) = 1 + O(ℏ ), xc ( 1 )  − 2 −1/2 2 ℏ2 1 q xcyc = 1 + O( ). yc These are rewritten, { 1/2 2 (1 − xc)yc = q + O(ℏ ), 1/2 2 1/2 2 q yc = (yc − 1) (yc − q ) + O(ℏ ).

2 2 Hence, putting xc = x0 + x1ℏ + O(ℏ ) and yc = y0 + y1ℏ + O(ℏ ) by Lemma 5.1, we have that    x y 1  1 − 1 + 1 = , x0 = 1 − , − y 1 x0 y0 2  0  1 ( 1) y = (y − 1)3, y + y = (y − 1)2 3y − . 0 0 1 2 0 0 1 2 Therefore, we obtain that 2 y0 y0 + 1 y1 = , x1 = − . 2(2y0 + 1) 2y0(2y0 + 1)

30 Further, in order to show Lemmas 5.2, 5.3 and 5.4 below, we calculate the expansion (48) concretely, as follows. Putting tˇ= t − tc, we expand φ(1 − t)/N at tc as follows, φ(1−t) φ(1−t ) φ′(1−t ) 1 φ′′(1−t ) 1 φ(3)(1−t ) 1 φ(4)(1−t ) = c − c tˇ+ c tˇ2− c tˇ3+ c tˇ4+··· , N N N 2 N 6 N 24 N

(k) where we can calculate φ (1 − tc)/N concretely by Proposition A.1. Hence, putting tˇ= t − tc ands ˇ = s − sc, we can show by concrete calculation that V (t, s) is expanded at a critical point (tc, sc) in the following form,

V (t, s) = V (tc, sc)+(degree 2 part)+(degree 3 part)+(degree 4 part)+(degree ≥ 5 part), (51) where ( ) √ tˇ2 1 sˇ2 1 + y (tˇ+s ˇ)2 (degree 2 part) = 2π −1 + c − + O(ℏ2), (52) 2 1 − x 2 1 − y 2 c ( c ) √ tˇ3 sˇ3 (degree 3 part) = (2π −1)2 (y − 1) · y + · 2 + O(ℏ), (53) 0 6 0 6 ( ) √ tˇ4 sˇ4 (degree 4 part) = (2π −1)3 (x2 + x )y3 − 2(y + 1) + O(ℏ). (54) 24 0 0 0 24 0

Now, we show lemmas used in the proof of Theorem 1.1. Lemma 5.2. Under the notation in the proof of Theorem 1.1, ( ) √ ( ) ( ) Nς −π −1/4 1/2 2 exp NV (tc, sc) = e e 1 − y0 1 + (C1 + C2)ℏ + O(ℏ ) ,

where we put 1 1 C = (−16y2 + 33y + 9),C = (2y2 − 5y ). 1 184 0 0 2 24 0 0 √ − · ··· ··· 2π 1 tc Proof. We put√ tc = t0 +t1/N + and sc = s0 +s1/N + . Recalling that xc = e 2π −1 ·sc and yc = e , we have that

x1 y0 + 1 y1 y0 t1 = = − , s1 = = . x0 2(y0 − 1)(2y0 + 1) y0 2(2y0 + 1)

Putting tˇ = −t1/N ands ˇ = −s1/N in the degree 2 part (52) of the expansion (51), we have that ( ) √ t2 1 s2 1 + y (t + s )2 − 1 1 0 − 1 1 ℏ3 V (t0, s0) = V (tc, sc) + 2π 1 2 + 2 2 + O( ). 2N 1 − x0 2N 1 − y0 2N Hence, 2 NV (tc, sc) = NV (t0, s0) + C1ℏ + O(ℏ ),

31 where we put ( ) 1 1 1 + y 1 − 2 2 0 − 2 − 2 C1 = t1 + s1 (t1 + s1) = ( 16y0 + 33y0 + 9). 2 1 − x0 1 − y0 184 Hence, ( ) ( ) ( ) 2 exp NV (tc, sc) = exp NV (t0, s0) · 1 + C1ℏ + O(ℏ ) . (55)

Further, we calculate V (t0, s0), as follows. From the deﬁnition of V (t, s), we have that 1 ( ) 1 π2 V (t0, s0) = − φ(1 − t0) − 2 φ(1 − s0) − √ N 2π −1 6 ( ) √ √ √ 1( 1 ) 1 1 1 3π −1 π −1 − 2π −1 t + s − 2 + s2 − t − s + − + . 2 0 0 2N 2 0 2 0 0 6 4N 4N 2 ˆ Since this goes to V (t0, s0) as N → ∞, we have that ( ( ) ( ) 2 ) ˆ 1 1 1 π V (t0, s0) = √ − Li2 − 2 Li2 − 2π −1 x y 6 ( 0 0 ) √ 1 1 1 1 − − 2 2 − − 2π 1 (t0 + s0) + s0 t0 s0 + ( 2 2 2 6 ) 1 ( 1 ) ( 1 ) ( 1 ) π2 = √ Li2(x0y0) + Li2 + Li2(y0) − Li2 − Li2 − = ς. 2π −1 x0y0 y0 x0 6 Hence, by Proposition A.1, √ √ √ − − ( ) ( ) − t0 + s0 − 3π 1 2π 1 1 1 1 V (t0, s0) = ς + 2π 1 + 2 + + O 3 . 2N 4N N 24(x0 −1) 12(y0 −1) N Therefore, ( ) √ ( ) ( ) Nς − 3 π −1 1/2 2 exp NV (t0, s0) = e e 4 x0y0 1 + C2ℏ + O(ℏ ) √ ( ) ( ) Nς −π −1/4 1/2 2 = e e 1 − y0 1 + C2ℏ + O(ℏ ) , where we put 1 1 1 2 − C2 = + = (2y0 5y0). 24(x0 − 1) 12(y0 − 1) 24 Hence, from the above formula and (55), we obtain the required formula. Lemma 5.3. Under the notation in the proof of Theorem 1.1, ( ) − 2 − 2 − − − ℏ ℏ2 VttVss Vts = 4π ( 2y0 1) 1 2C3 + O( ) , ( ) ( ) 2 −1/2 1 −1/2 2 (VttVss − V ) = √ − 2y0 − 1 1 + C3ℏ + O(ℏ ) , ts 2π −1 where we put 1 ( ) C = 8y2 − 17y + 9 . 3 46 0 0

32 Proof. From the degree 2 part (52) of the expansion (51), we have that ( ) ∂2V √ 1 √ x − − ℏ2 − · c ℏ2 Vtt = 2 (tc, sc) = 2π 1 1 + O( ) = 2π 1 + O( ), ∂t 1 − xc 1 − xc 2 √ ∂ V 2 Vts = (tc, sc) = −2π −1 + O(ℏ ), ∂t ∂s ( ) ∂2V √ 1 + y √ 2y − c − ℏ2 − · c ℏ2 Vss = 2 (tc, sc) = 2π 1 1 + O( ) = 2π 1 + O( ) ∂s 1 − yc 1 − yc Hence, ( ) x 2y − 2 − 2 c · c − ℏ2 VttVss Vts = 4π 1 + O( ) 1 − xc 1 − yc ( ) 2q−1/2x y2 = −4π2 c c − 1 + O(ℏ2) 1 − yc ( ) 1 − q−1/2y − 2 − · c − ℏ2 = 4π 2yc − 1 + O( ) ( 1 yc ) ( y ) = −4π2 − y 2 + 0 ℏ − 1 + O(ℏ2) c 1 − y ( 0 ) y2(y + 2) = −4π2(−2y − 1) 1 − 0 0 ℏ + O(ℏ2) 0 − 2 ( (y0 ) 1)(2y0 + 1) 2 2 = −4π (−2y0 − 1) 1 − 2C3ℏ + O(ℏ ), where y2(y + 2) 1 ( ) 0 0 2 − C3 = 2 = 8y0 17y0 + 9 . 2(y0 − 1)(2y0 + 1) 46 Therefore, we obtain the required formulas. Lemma 5.4. Under the notation in the proof of Theorem 1.1,

λ1 = C4 + C5 , where we put 1 1 C = (2y2 + 15y ),C = − (197y2 + 127y + 227). 4 92 0 0 5 3174 0 0

Proof. We show the lemma by using Proposition 3.5. As shown in Proposition 3.5, λ1 consists of two contributions from the degree 3 part and the degree 4 part of the expansion (51). We calculate the contribution from the degree 4 part of the expansion (51), as follows. As in (54), the degree 4 part of this expansion is given by ( ) √ tˇ4 sˇ4 (2π −1)3 (x2 + x )y3 − 2(y + 1) + O(ℏ). (56) 24 0 0 0 24 0

33 As explained in Proposition 3.5, the contribution from tˇ4 is calculated as ( ( ) ( ) ) ( ) T ∂ 4 1 u − u N · exp − 1 A 1 1 ∂u1 4N u2 u2 u =u =0 ( ( ) ( ) 1 ( 2 ) ) ( ) T −1 2 1 ∂ 4 1 u V V u = N · − 1 tt ts 1 2 ∂u1 2N u2 Vts Vss u2 1 ( ∂ ) ( ) = 4 V u2 − 2V u u + V u2 2 − 2 2 ss 1 ts 1 2 tt 2 8N (VttVss Vts) ∂u1 1 = · 4! · V 2 − 2 2 ss 8N (VttVss Vts) 4! ( 2y ) = √ 0 2 + O(ℏ), 2 2 − 8N (2π −1) (2y0 + 1) 1 y0 by using Lemma 5.3 and formulas in the proof of Lemma 5.3. Similarly, the contribution froms ˇ4 is calculated as 1 ( ∂ ) ( ) 4 V u2 − 2V u u + V u2 2 − 2 2 ss 1 ts 1 2 tt 2 8N (VttVss Vts) ∂u2 1 = · 4! · V 2 − 2 2 tt 8N (VttVss Vts) 4! ( x ) = √ 0 2 + O(ℏ). 2 2 − 8N (2π −1) (2y0 + 1) 1 x0 Hence, the contribution from (56) is equal to √ ( ) 2π −1 ( 2y ) ( x ) 2 3 0 2 − 0 2 ℏ 2 (x0 + x0)y0 2(y0 + 1) = C4 , 8N (2y0 + 1) 1 − y0 1 − x0 where we put ( ) 1 ( 2y ) ( x ) 1 2 3 0 2 − 0 2 2 C4 = 2 (x0 + x0)y0 2(y0 + 1) = (2y0 + 15y0). 8(2y0 + 1) 1 − y0 1 − x0 92 We calculate the contribution from the degree 3 part of the expansion (51), as follows. As in (53), the degree 3 part of this expansion is given by ( ) √ tˇ3 sˇ3 (2π −1)2 (y − 1) · y + · 2 + O(ℏ), 0 6 0 6 and, hence, the corresponding degree 6 part is given by ( ) 1 √ tˇ3 sˇ3 2 (2π −1)4 (y − 1)2 · y + · 2 + O(ℏ) 2 0 6 0 6 ( ) 1 √ tˇ6 sˇ6 tˇ3sˇ3 = (2π −1)4 (y − 1)2 · y2 + · 4 + · 4y + O(ℏ). (57) 2 0 62 0 62 62 0

34 The contribution from tˇ6 is calculated as ( ( ) ( ) ) ( ) T ∂ 6 1 u − u N 2 · exp − 1 A 1 1 ∂u1 4N u2 u2 u =u =0 ( ( ) ( ) 1 ( 2 ) ) ( ) T −1 3 1 ∂ 6 1 u V V u = N 2 · − 1 tt ts 1 6 ∂u1 2N u2 Vts Vss u2 −1 ( ∂ ) ( ) = 6 V u2 − 2V u u + V u2 3 − 2 3 ss 1 ts 1 2 tt 2 48N (VttVss Vts) ∂u1 −1 = · 6! · V 3 − 2 3 ss 48N (VttVss Vts) 6! ( 2y ) = √ 0 3 + O(ℏ), 3 3 − 48N (2π −1) (2y0 + 1) 1 y0 by using Lemma 5.3 and formulas in the proof of Lemma 5.3. Similarly, the contribution froms ˇ6 is calculated as −1 ( ∂ ) ( ) 6 V u2 − 2V u u + V u2 3 − 2 3 ss 1 ts 1 2 tt 2 48N (VttVss Vts) ∂u2 −1 = · 6! · V 3 − 2 3 tt 48N (VttVss Vts) 6! ( x ) = √ 0 3 + O(ℏ). 3 3 − 48N (2π −1) (2y0 + 1) 1 x0 Similarly, the contribution from tˇ3sˇ3 is calculated as −1 ( ∂ ) ( ∂ ) ( ) 3 3 V u2 − 2V u u + V u2 3 48N (V V − V 2)3 ∂u ∂u ss 1 ts 1 2 tt 2 tt ss ts 1 ( 2 ) −1 2 3 = · 6! · (−V )3 + V V (−V ) + O(ℏ) 48N (V V − V 2)3 5 ts 5 tt ss ts tt ss ts ( ) 6! 2 3 x 2y = √ + 0 0 , 3 3 − − 48N (2π −1) (2y0 + 1) 5 5 1 x0 1 y0 by using Lemma 5.3 and formulas in the proof of Lemma 5.3. Hence, the contribution from (57) is equal to √ ( ) 5 · 2π −1 (y − 1)2 ( 2y ) ( x ) (2 3 x 2y ) 0 2 0 3 0 3 0 0 ℏ 3 y0 + 4 + 4y0 + = C5 24N (2y0 + 1) 1 − y0 1 − x0 5 5 1 − x0 1 − y0 where we put − 2 ( ( ) ( ) ( )) 5(y0 1) 2 2y0 3 x0 3 2 3 x0 2y0 C5 = 3 y0 + 4 + 4y0 + 24(2y0 + 1) 1 − y0 1 − x0 5 5 1 − x0 1 − y0 1 = − (197y2 + 127y + 227). 3174 0 0 Therefore, we obtain the required formula of the lemma.

35 5.3 Verifying the assumption of the saddle point method In this section, in Lemma 5.7, we verify the assumption of the saddle point method (Proposition 3.5, see also Remark 3.6 and Appendix B) when we apply Proposition 3.5 and Remark 3.6 to (46). The arguments of this section are due to Yokota [44]. Let V (t, s) and Vˆ (t, s) be as in Section 2.2. As shown in Appendix B, V (t, s) uniformly converges to Vˆ (t, s) on ∆′ as N → ∞. Hence, as mentioned in Remarks 3.3 and 3.6 and Appendix B, the saddle point method∫ of the problem can be reduced to the saddle point method of an integral of the form ϕ(t, s) eN Vˆ (t,s)dt ds. Therefore, we verify the assumption of the saddle point method for Vˆ (t, s). We recall that the diﬀerentials of Vˆ (t, s) are given in (47). In order to show Lemma 5.7 below, we calculate the behavior of the following function, √ √ ˆ − − − ft,s(δ1, δ2) = Re V (t + δ1 1, s + δ2 1) ςR . The diﬀerentials of this function are given by ( ) ∂ √ ∂ √ √ f (δ , δ ) = Re −1 Vˆ (t + δ −1, s + δ −1) ∂δ t,s 1 2 ∂t 1 2 1 ( ) ( 1 ) √ ( 1) = −Im − log 1 − − 2π −1 t + s − x 2 ( 1 ) ( 1) = Arg 1 − + 2π t + s − , (58) ( x 2 ) ∂ √ ∂ √ √ f (δ , δ ) = Re −1 Vˆ (t + δ −1, s + δ −1) ∂δ t,s 1 2 ∂s 1 2 2 ( ) ( 1) √ = −Im − 2 log 1 − − 2π −1 (t + 2s − 1) y ( ) ( 1) (1 1) = 2 Arg 1 − + 2π t + s − , (59) y 2 2 √ √ √ √ − − − − where x = e2π 1 (t+δ1 1) and y = e2π 1 (s+δ2 1). ′ Lemma 5.5 (Yokota [44]). Fixing (t, s) ∈ ∆ and δ2 ∈ R, we regard ft,s(X, δ2) as a function of X ∈ R. ≥ 1 ∈ R (1) If t + s 2 , then ft,s(X, δ2) is monotonically increasing for X . 1 (2) If t + s < 2 , then ft,s(X, δ2) has a unique minimal point at X = g1(t, s), where 1 sin 2π(t + s) g (t, s) = log , 1 2π sin 2πs

i.e., ft,s(X, δ2) is monotonically decreasing for X < g1(t, s), and is monotonically increasing for X > g1(t, s). √ √ √ 2π −1 (t+X −1) 2πX −2π −1 t − 1 Proof. We put x = e . Then, 1/x = e e . We put θ = Arg (1 x ) 1 in this proof. Since t < 2 , θ is in the following range, (1 ) 0 < θ < 2π − t . 2

36 ≥ 1 When t + s 2 , we show the lemma, as follows. By (58), ∂ ( 1) f (X, δ ) = θ + 2π t + s − > 0. ∂X t,s 2 2

Therefore, ft,s(X, δ2) is monotonically increasing, and (1) holds. When t + s < 1 , we show the lemma, as follows. In this case, by (58), 2   1 − − > 0 if θ > 2π( 2 t s), ∂ 1 ft,s(X, δ2) = 0 if θ = 2π( − t − s), ∂X  2 1 − − < 0 if θ < 2π( 2 t s). Further, θ and X are related as shown in the following picture.

1 0 1 2πt θ

e2πX 1/x Hence, X is monotonically increasing as a function of θ, and they satisfy that e2πX 1 = . sin θ sin(π − 2πt − θ) This is rewritten, 1 sin θ X = log . 2π sin(π − 2πt − θ) Therefore,  > 0 if X > g (t, s), ∂ 1 ft,s(X, δ2) = 0 if X = g (t, s), ∂X  1 < 0 if X < g1(t, s), where we put 1 sin 2π( 1 − t − s) 1 sin 2π(t + s) g (t, s) = log 2 = log . 1 2π sin 2πs 2π sin 2πs Hence, (2) holds. ′ Lemma 5.6 (Yokota [44]). Fixing (t, s) ∈ ∆ and δ1 ∈ R, we regard ft,s(δ1,Y ) as a function of Y ∈ R. (1) If t + 2s ≥ 1, then ft,s(δ1,Y ) is monotonically increasing for Y ∈ R. (2) If t + 2s < 1, then ft,s(δ1,Y ) has a unique minimal point at Y = g2(t, s), where 1 sin π(t + 2s) g (t, s) = log , 2 2π sin πt

i.e., ft,s(δ1,Y ) is monotonically decreasing for Y < g2(t, s), and is monotonically increasing for Y > g2(t, s).

37 √ √ √ 2π −1 (s+Y −1) 2πY −2π −1 s − 1 Proof. We put y = e . Then, 1/y = e e . We put θ = Arg (1 y ) 1 in this proof. Since s < 2 , θ is in the following range, (1 ) 0 < θ < 2π − s . 2 When t + 2s ≥ 1, we show the lemma, as follows. By (59), ( ) ∂ ( ) f (δ ,Y ) = 2 θ + π t + 2s − 1 > 0. ∂Y t,s 1

Hence, ft,s(δ1,Y ) is monotonically increasing, and (1) holds. When t + 2s < 1, we show the lemma, as follows. In this case, by (59),   1 − 1 − > 0 if θ > 2π( 2 2 t s), ∂ 1 1 ft,s(δ1,Y ) = 0 if θ = 2π( − t − s), ∂Y  2 2 1 − 1 − < 0 if θ < 2π( 2 2 t s). Further, similarly as in the proof of Lemma 5.5, θ and Y are related by 1 sin θ Y = log . 2π sin(π − 2πs − θ) Since Y is monotonically increasing as a function of θ,  > 0 if Y > g (t, s), ∂ 2 ft,s(δ1,Y ) = 0 if Y = g (t, s), ∂Y  2 < 0 if Y < g2(t, s), where we put 1 sin 2π( 1 − 1 t − s) 1 sin π(t + 2s) g (t, s) = log ( 2 2 ) = log . 2 2π − − 1 − 1 − 2π sin πt sin π 2πs 2π( 2 2 t s) Hence, we obtain the lemma. Lemma 5.7 (Yokota [44]). When we apply Proposition 3.5 (saddle point method) to (46), the assumption of Proposition 3.5 holds. ′ ≤ ≤ ′ ′ ′ Proof. We show that there exists a homotopy ∆δ (0 δ 1) between ∆0 = ∆ and ∆1 such that (t , s ) ∈ ∆′ , (60) c c 1 { } ∆′ − {(t , s )} ⊂ (t, s) ∈ C2 Re Vˆ (t, s) < ς , (61) 1 c{ c } R ′ ⊂ ∈ C2 ˆ ∂∆δ (t, s) Re V (t, s) < ςR . (62) For a suﬃciently large R > 0, we put { max {−R, g (t, s)} if t + s < 1 , gˆ (t, s) = 1 2 1 − ≥ 1 R if t + s 2 ,

38 { max {−R, g (t, s)} if t + 2s < 1, gˆ (t, s) = 2 2 −R if t + 2s ≥ 1.

→ −∞ → 1 We note that, since g1(t, s) as t+s 2 ,g ˆ1(t, s) is continuous, and similarly, since g2(t, s) → −∞ as t + 2s → 1,g ˆ2(t, s) is continuous. We put {( √ √ ) } ′ · − · − ∈ C2 ∈ ′ ∆δ = t + δ gˆ1(t, s) 1, s + δ gˆ2(t, s) 1 (t, s) ∆ . We show (62), as follows. From the definition of ∆′, { } ′ ⊂ ∈ C2 ˆ ∂∆ (t, s) Re V (t, s) < ςR . Further, by Lemmas 5.5 and 5.6, ( √ √ ) ˆ ˆ Re V t + δ · gˆ1(t, s) −1, s + δ · gˆ2(t, s) −1 ≤ V (t, s) for any δ ∈ [0, 1] and any (t, s) ∈ ∆′. Hence, (62) holds. We show (60) and (61), as follows. Consider the following functions ( √ √ ) F (t, s, X, Y ) = Re Vˆ t + X −1, s + Y −1 , ( ) h(t, s) = F t, s, gˆ1(t, s), gˆ2(t, s) . When t + 2s ≥ 1, −h(t, s) is sufficiently large (because we let R be sufficiently large), and (61) holds in this case. When t + 2s < 1, it is shown from the definitions of g1(t, s) and g (t, s) that ∂F = 0 at X = g (t, s) and ∂F = 0 at Y = g (t, s). Hence, Im ∂Vˆ = 2 ( ∂X √ 1 √ ∂Y) 2 ∂t ∂Vˆ − − ∂h ∂Vˆ ∂h ∂Vˆ Im(∂s = 0 at √t + g1(t, s) 1,√ s + g)2(t, s) 1 . Further, ∂t = Re ∂t and ∂s = Re ∂s − − at( t + g1(t,√ s) 1, s + g2(t,√ s) ) 1 . Therefore, when (t, s) is a critial point of h(t, s), ˆ t + g1(t, s) −1, s + g2(t, s) −1 is a critical point of V . It follows that h(t, s) has a unique maximal point at (t, s) = (Re tc, Re sc). Therefore, (60) and (61) hold.

5.4 Verifying the assumption of the Poisson summation formula In this section, in Lemma 5.8, we verify the assumption of the Poisson summation formula (Proposition 4.6, see also Remark 4.8 and Appendix B) when we apply Proposition 4.6 and Remark 4.8 to (14). As in the previous section, we consider Vˆ (t, s) instead of V (t, s) in this section. √ ∈ ′ ˆ 2π −1 ·t We calculate√ a critical point (tr, sr) ∆ of Re V (t, s), as follows. Putting x = e and y = e2π −1 ·s, we have that ( ) ∂ ( 1 ) Re Vˆ (t, s) = Re − log 1 − , ∂t ( x ) ∂ ( 1) Re Vˆ (t, s) = Re − 2 log 1 − , ∂s y √ √ 2π −1 ·tr 2π −1 ·sr since t, s ∈ R. Hence, putting xr = e and yr = e ,

xr − 1 = yr − 1 = 1,

39 √ 2π −1/3 noting that |xr| = |yr| = 1. Therefore, xr = yr = e , and Re V (t, s) has a unique 1 1 ′ maximal point ( 6 , 6 ) on ∆ . Its maximal value is given by ( ) ( √ √ ) ˆ 1 1 1 π −1/3 −π −1/3 Re V , − ς = Re √ Li2(e ) − 2 Li2(e ) − ς 6 6 R 2π −1 R = 0.03448931080... . Hence, ˆ − ≤ Re V (t, s) ςR 0.03448931080... , (63) for any (t, s) ∈ ∆′. Lemma 5.8. When we apply Proposition 4.6 to (14), the assumptions of Proposition 4.6 hold. Proof. We verify the assumptions (41) and (42) for i = 1 in Lemmas 5.9 and 5.10 below, and verify the assumptions for i = 2 in Lemmas 5.11 and 5.12 below. The other assumptions of Proposition 4.6 can be verified easily. Lemma 5.9. The assumption (41) holds for i = 1. Proof. As for the assumption (41) for i = 1, we show that ∂∆′ is null-homotopic in { √ √ } − ∈ C2 ∈ ′ ≥ − (t + δ 1, s) (t, s) ∆ , δ 0, Re V (t + δ 1, s) < ςR + 2πδ . To show it, we show that the following disk bounds ∂∆′ in the above domain, { √ } { √ } 2 ′ 2 ′ (t + δ0 −1, s) ∈ C (t, s) ∈ ∆ ∪ (t + δ −1, s) ∈ C (t, s) ∈ ∂∆ , δ ∈ [0, δ0] . We put √ − − − Ft,s(δ) = Re V (t + δ 1, s) ςR 2πδ in this proof. Then, it is sufficient to show that ′ Ft,s(δ0) < 0 for any (t, s) ∈ ∆ , and (64) ′ Ft,s(δ) < 0 for any (t, s) ∈ ∂∆ and δ ∈ [0, δ0], (65) for some δ0 > 0. To show these, we estimate the differential of Ft,s(δ), as follows. The differential of Ft,s(δ) is given by (√ √ ) d ∂ ˆ Ft,s(δ) = Re −1 V (t + δ −1, s) − 2π dδ ( ∂t ) ( 1 ) √ ( 1) = −Im − log 1 − − 2π −1 t + s − − 2π x 2 ( 1 ) ( 3) = Arg 1 − + 2π t + s − , x 2 √ √ 2π −1(t+δ −1) ≤ ≤ − 1 where we put x = e . Since 0.04 t 0.4, it is shown that Arg (1 x ) is in the following range, ( 1 ) 0 < Arg 1 − < π. x

40 Therefore, since t + s ≤ 0.6, d (1 3) F (δ) < 2π + 0.6 − = −2π · 0.4. dδ t,s 2 2 We show (64), as follows. We have that ∫ δ0 d Ft,s(δ0) = Ft,s(0) + Ft,s(δ) dδ < Ft,s(0) − 2π · 0.4 · δ0 . 0 dδ Further, by (63), ˆ − ≤ Ft,s(0) = Re V (t, s) ςR 0.03448931080... .

Hence, (64) is satisfied for a sufficiently large δ0. ′ We show (65), as follows. From the definition of ∆ , we have that Ft,s(0) < 0 for any ∈ ′ d (t, s) ∂∆ . Since dδ Ft,s(δ) < 0 as shown above, it is shown similarly as above that Ft,s(δ) < 0 for any δ ≥ 0. Hence, (65) is satisfied. Lemma 5.10. The assumption (42) holds for i = 1. Proof. We put √ − − − − Ft,s(δ) = Re V (t δ 1, s) ςR 2πδ in this proof. Similarly as the proof of Lemma 5.9, it is sufficient to show that there exists ε > 0 such that d F (δ) < −ε, (66) dδ t,s for any (t, s) ∈ ∆′. We calculate this as ( √ √ ) d ∂ ˆ Ft,s(δ) = Re − −1 V (t − δ −1, s) − 2π dδ ( ∂t ) ( 1 ) √ ( 1) = Im − log 1 − − 2π −1 t + s − − 2π x 2 ( 1 ) ( 1) = −Arg 1 − − 2π t + s + , x 2 √ √ 2π −1(t−δ −1) ≤ ≤ − 1 where we put x = e . Since 0.04 t 0.4, it is shown that Arg (1 x ) is in the following range, ( 1 ) 0 < Arg 1 − < π . x Therefore, since t + s ≥ 0.09, d ( 1) F (δ) < −2π 0.09 + = −2π · 0.59, dδ t,s 2 and (66) is satisfied. Lemma 5.11. The assumption (41) holds for i = 2.

41 Proof. We put √ − − − Ft,s(δ) = Re V (t, s + δ 1) ςR 2πδ in this proof. Similarly as the proof of Lemma 5.9, it is sufficient to show that there exists ε > 0 such that d F (δ) < −ε, (67) dδ t,s for any (t, s) ∈ ∆′. To show these, we calculate the differential of Ft,s(δ) as ( ) d √ ∂ √ Ft,s(δ) = Re −1 V (t, s + δ −1) − 2π dδ ( ∂s ) ( 1) √ ( ) = −Im − 2 log 1 − − 2π −1 t + 2s − 1 − 2π y ( 1) ( ) = 2 Arg 1 − + 2π t + 2s − 2 , y √ √ 2π −1(s+δ −1) ≤ ≤ − 1 where we put y = e . Since 0.05 s 0.4, it is shown that Arg (1 y ) is in the following range, ( 1) (1 ) 0 < Arg 1 − < 2π − s . y 2 Therefore, since t ≤ 0.4, d (1 ) ( ) F (δ) < 2 · 2π − s + 2π t + 2s − 2 = 2π(t − 1) ≤ −2π · 0.6, dδ t,s 2 and (67) is satisfied. Lemma 5.12. The assumption (42) holds for i = 2. Proof. We put √ − − − − Ft,s(δ) = Re V (t, s δ 1) ςR 2πδ in this proof. Similarly as the proof of Lemma 5.9, it is sufficient to show that there exists ε > 0 such that d F (δ) < −ε, (68) dδ t,s for any (t, s) ∈ ∆′. We calculate this as ( ) d √ ∂ √ Ft,s(δ) = Re − −1 V (t, s − δ −1) − 2π dδ ( ∂s ) ( 1) √ ( ) = Im − 2 log 1 − − 2π −1 t + 2s − 1 − 2π y ( 1) ( ) = −2 Arg 1 − − 2π t + 2s , y

42 √ √ 2π −1(s−δ −1) ≤ ≤ − 1 where we put y = e . Since 0.05 s 0.4, it is shown that Arg (1 y ) is in the following range, ( 1) 0 < Arg 1 − < π . y Therefore, since t + s ≥ 0.09 and s ≥ 0.05, d ( ) F (δ) < −2π 0.09 + 0.05 = −2π · 0.14, dδ t,s and (68) is satisﬁed.

A Properties of φ(t)

In this appendix, we√ review some basic properties of φ(t). We put ℏ = 2π −1/N, and put ( ) ( ) ∑ d 2k−2 z Φ (z) = Li z + ℏ2kc · z , d 2 2k dz 1 − z 1≤k≤d

where we define c2k by y/2 ∑ = c y2k. sinh(y/2) 2k k≥0 Proposition A.1. We fix any sufficiently small δ > 0 and any M > 0. Let d be any non-negative integer. Then, in the domain { } t ∈ C δ ≤ Re t ≤ 1 − δ, |Im t| ≤ M , (69)

φ(t) and φ(k)(t) are presented by √ ( ) N 2π −1 t 1 φ(t) = √ Φd(e ) + O , (70) 2π −1 N 2d+1 ( ) √ ( ) (k) N d k 2π −1 t 1 φ (t) = √ Φd(e ) + O , (71) 2π −1 dt N 2d+1 for each k > 0, where O(1/N 2d+1) means the error term whose absolute value is bounded 2d+1 by C/N for some C > 0, which is independent of t (but√ possibly dependent on δ). 1 √1 2π −1 t In particular, φ(t) uniformly converges to Li2(e ) in the domain (69), and N √ 2π −1 1 ′ − − 2π −1 t N φ (t) uniformly converges to log(1 e ) in the domain (69).

Proof. We show (70). We have that ∫ ∫ ∞ (2t−1)x ∞ (2t−1)x ∑ ( ) Ne · x/N Ne 2x 2k φ(t) = 2 dx = 2 c2k dx. (72) −∞ 4x sinh x sinh(x/N) −∞ 4x sinh x N k≥0

43 We put the “k =0” part of (72) to be ∫ ∞ Ne(2t−1)x f(t) = 2 dx. −∞ 4x sinh x To calculate f(t), we consider the following contour, [ ] { − (m + 1 )π, −1 ∪ z ∈ C |z| = 1, Im z ≥ 0} [ 2 ] { } ∪ 1 ∪ ∈ C | | 1 ≥ 1, (m + 2 )π z z = m + 2 , Im z 0 . √ Then, the integrand has poles at nπ −1 (n = 1, 2, ··· , m) in the region bounded by the contour. Hence,

∑m √ (2t−1)x − Ne f(t) = lim 2π 1 Res√ 2 m→∞ x=nπ −1 4x sinh x n=1 ∞ √ √ ∑ −N (e2π −1 t)n N √ = 2π −1 = √ Li (e2π −1 t). 4π2 n2 − 2 n=1 2π 1 Further, for general k > 0, the “k” part of (72) is calculated as ( ) √ ( ) c2k (2k) N c2k d 2k 2π −1 t N 2k d 2k−2 z f (t) = √ · Li2(e ) = √ · ℏ c2k z . N 2k 2π −1 N 2k dt 2π −1 dz 1 − z Therefore, we can put √ N 2π −1 t φ(t) = √ Φd(e ) + Rd, 2π −1

where Rd is the error term. We estimate it, as follows. Since |(y/2)/ sinh(y/2)| is bounded for y ∈ R, there exists M0 > 0 such that

1 ( y/2 ∑d ) − c y2k ≤ M y2d+2 sinh y/2 2k 0 k=0 for any y ∈ R. Hence, putting y = 2x/N, √ N 2π −1 t |Rd| = φ(t) − √ Φd(e ) 2π −1 ∫ ( d ) ∞ Ne(2t−1)x x/N ∑ (2x) − 2k = 2 c2k dx −∞ 4x sinh x sinh(x/N) N k=0 ∫ ∞ Ne(2t−1)x 2x ≤ · 2d+2 2 M0 dx −∞ 4x sinh x N ( ∫ ∫ ∫ ) 22dM ∞ e(2Re t−1)x −1 e(2Re t−1)x eRe ((2t−1)z) = 0 x2d dx + x2d dx + |dz| . 2d+1 | | | | | | | | N 1 sinh x −∞ sinh x z =1 sinh z Im z ≥ 0 (73)

44 −2δx 2d For x ≥ 1, the integrand of (73) is bounded by M1e x for some M1 > 0. For x ≤ −1, 2δx 2d the integrand of (73) is bounded by M2e x for some M2 > 0. For |z| = 1, the integrand of (73) is bounded by a constant. Hence, the values of the integrals of (73) is bounded by 2d+1 a constant. Therefore, |Rd| is bounded by C/N for some C > 0, which is independent of t. Hence, we obtain (70). We obtain (71) by similar arguments for φ(k)(t). Lemma A.2. For any t ∈ C with 0 < Re t < 1, ( ) √ N ( 1) 1 φ(t) + φ(1 − t) = 2π −1 − t2 − t + + . 2 6 24N Proof. By definition, ∫ e(2t−1)x + e(1−2t)x φ(t) + φ(1 − t) = dx. γ 4x sinh x sinh(x/N) Since this integrand is an odd function, if the contour was R, the integral would vanish. In this case, since the contour γ avoids the pole at x = 0, the contribution from the residue at this pole survives. By expanding the numerator and the denominator of the integrand as power series of x concretely, we can calculate this residue, and obtain the required formula. Lemma A.3 (Kashaev). √ √ ( 1 ) N π2 1 π −1 π −1 φ = √ + log N + − , − 2N 2π 1 6 2 4 √ 12N √ ( 1 ) N π2 1 π −1 π −1 φ 1 − = √ − log N + − . 2N 2π −1 6 2 4 12N Proof. It is sufficient to show that ( ) ( 1 ) ( 1 ) √ N 1 1 φ + φ 1 − = 2π −1 − + − , (74) 2N 2N 12 4 12N ( 1 ) ( 1 ) φ − φ 1 − = log N. (75) 2N 2N We obtain (74) from Lemma A.2 by putting t = 1/2N. We show (75), as follows. By (3) and (8), ( ) ( 1 ) ( 1 ) exp φ − φ 1 − = N. (76) 2N 2N Further, from the definition of φ(t), ∫ ( 1 ) ( 1 ) e(1−1/N)x − e(1/N−1)x φ − φ 1 − = dx. 2N 2N γ 4x sinh x sinh(x/N) By a similar argument as in the proof of Lemma A.2, we can see that the residue of this integrand at 0 vanishes. Hence, the value of this integral is real, and we obtain (75) from (76).

45 B Convergence of V (t, s) to Vˆ (t, s)

When we apply the saddle point method and the Poisson summation formula to V (t, s), it is necessary to note how V (t, s) converges to Vˆ (t, s) as N → ∞. In this appendix, we verify that this convergence is suitable in those applications. We recall that in Section 2.2 we put V (t, s) and Vˆ (t, s) by 1 ( ) 1 π2 V (t, s) = − φ(1 − t) − 2 φ(1 − s) − √ N 2π −1 6 ( ) √ √ √ 1( 1 ) 1 1 1 3π −1 π −1 − 2π −1 t + s − 2 + s2 − t − s + − + 2 2N 2 2 6 4N 4N 2 and ( √ √ 2 ) ˆ 1 −2π −1 t −2π −1 s π V (t, s) = √ − Li2(e ) − 2 Li2(e ) − 2π −1 6 ( ) √ 1 1 1 − 2π −1 t2 + s2 + ts − t − s + . 2 2 6 Lemma B.1. Let m be any non-negative integer. Then, in the domain ∆′′ of (12), V (t, s) is presented by the following form 1 1 1 1 V (t, s) = Vˆ (t, s) + V (t, s) + V (t, s) + ··· + V (t, s) + +R (t, s) 1 N 2 N 2 m N m m N m+1

where Vi(t, s)’s are holomorphic functions independent of N, and |Rm(t, s)| is bounded by a constant which is independent of N. In particular, V (t, s) uniformly converges to Vˆ (t, s) in the domain ∆′′.

Proof. We obtain the required presentation by applying Proposition A.1√ to V (t, s). In 1 √1 2π −1 t → particular, by Proposition A.1, N φ(t) uniformly converges to 2π −1 Li2(e ) as N ∞ in the domain ∆′′. Hence, we obtain the lemma. ′′ Lemma B.2. Let i and j be any non-negative integers. Then, in the domain ∆ , ∂i+j i j V (t, s) is bounded by a constant which is independent of N. ∂ t∂ s ∂i+j ∂i+j ˆ → ∞ Proof. By Proposition A.1, ∂it∂j s V (t, s) uniformly converges to ∂it∂j s V (t, s) as N ′′ ′′ ∂i+j ˆ ′′ in the domain ∆ . Since ∆ is compact, ∂it∂j s V (t, s) is bounded in ∆ . Hence, we obtain the lemma.

When we apply the saddle point method, we need to show that V (t, s) satisfies the condition of Remark 3.6, noting that Vˆ (t, s) satisfies the assumption of the saddle point method as shown in Section 5.3. In fact, we can show the condition of Remark 3.6 for V (t, s) by Lemma B.1. When we apply the Poisson summation formula, we need to show that V (t, s) satisfies the condition of Remark 4.8, noting that Vˆ (t, s) satisfies the assumption of the Poisson summation formula as shown in Section 5.4. In fact, we can show the condition of Remark 4.8 for V (t, s) by Lemmas B.1 and B.2.

46 C Critical points of V (t, s) and Vˆ (t, s)

In this appendix, we√ show a proof of Lemma 5.1. We put ℏ = 2π −1/N, and use the notation in Section 5.2. By Proposition A.1, we can put 1 √ φ′(1 − t) = − log(1 − e−2π −1 t) + r(t, ℏ) ℏ2 N for some smooth function r(t, ℏ) of t and ℏ. Then, from the deﬁnition of V (t, s), we have that ∂ ( 1 ) √ ( 1 1 ) V (t, s) = − log 1 − − 2π −1 t + s − − + r(t, ℏ) ℏ2, ∂t x 2 2N ∂ ( 1) √ ( 1 ) V (t, s) = −2 log 1 − − 2π −1 t + 2s − 1 − + 2 r(s, ℏ) ℏ2. ∂s y 2N Hence, a critical point of V (t, s) is a solution of  ( ) 1 2  1 − (−q−1/2xy) = er(t,ℏ) ℏ , x ( )  1 2  1 − 2q−1/2xy2 = e2 r(s,ℏ) ℏ . y This is rewritten as { (1 − x)y = q1/2er(t,ℏ) ℏ2 , x(y − 1)2 = q1/2e2 r(s,ℏ) ℏ2 . Further, by putting F (t, s, ℏ) = (1 − x)y − q1/2er(t,ℏ) ℏ2 , G(t, s, ℏ) = x(y − 1)2 − q1/2e2 r(s,ℏ) ℏ2 , the above system of equations is rewritten as { F (t, s, ℏ) = 0, (77) G(t, s, ℏ) = 0.

We note that (t0, s0, 0) is a solution of (77). ( ) Proof of Lemma 5.1. It is sufficient to show that there exists a smooth solution f(ℏ), g(ℏ), ℏ of (77) in a sufficiently small neighborhood of (t0, s0, 0). Hence, by the implicit function theorem, it is sufficient to show that ( ) ∂ ∂ F (t0, s0, 0) F (t0, s0, 0) det ∂t ∂s ≠ 0. (78) ∂ ∂ ∂t G(t0, s0, 0) ∂s G(t0, s0, 0) √ √ √ 2π −1 t d − d d − d Since x = e , dt = 2π 1 x dx . Similarly, ds = 2π 1 y dy . Therefore, from the definitions of F (t, s, ℏ) and G(t, s, ℏ), we have that ( ) 1 ∂ ∂ √ F (t0, s0, 0) = x (1 − x)y = −x0y0 , 2π −1 ∂t ∂x ℏ=0

47 ( ) 1 ∂ ∂ √ F (t0, s0, 0) = y (1 − x)y = (1 − x0)y0 , 2π −1 ∂s ∂y ℏ=0 ( ) 1 ∂ ∂ 2 2 √ G(t0, s0, 0) = x x(y − 1) = x0(y0 − 1) , 2π −1 ∂t ∂x ℏ=0 ( ) 1 ∂ ∂ 2 √ G(t0, s0, 0) = y x(y − 1) = 2x0y0(y0 − 1). 2π −1 ∂s ∂y ℏ=0 Hence, the determinant of (78) is calculated as ( ) 2 −x0y0 (1 − x0)y0 2 − 4π · det 2 = 4π · x0y0(y0 − 1)(x0y0 + x0 + y0 − 1) x0(y0 − 1) 2x0y0(y0 − 1) 2 3 2 = 4π · (2y0 + 1)(y0 − 1) /y0 = 4π · (2y0 + 1) ≠ 0, 3 since x0 = 1 − 1/y0 and (y0 − 1) = y0. Therefore, we obtain (78), completing the proof of the lemma.

D Proof of Lemma 2.1

In this appendix, we give a proof of Lemma 2.1. Proof of Lemma 2.1. We put ( √ ) 1 2π −1 t Λ(t) = Re √ Li2(e ) 2π −1 for t ∈ R. We note that this function has period 1, and Λ(−t) = −Λ(t). Since ( ) Λ′(t) = − log 2 sin πt for 0 < t < 1, the behavior of Λ(t) is as follows. 1 1 5 t 0 ··· 6 ··· 2 ··· 6 ··· 1 1 → → − 1 Λ(t) 0 → Λ( 6 ) 0 Λ( 6 ) → 0 1 Here, Λ( 6 ) = 0.161533... . See Figure 6 for the graph of Λ(t).

0.15 0.10 0.05 0.2 0.4 0.6 0.8 1.0

− 0.05 − 0.10 − 0.15

Figure 6: The graph of Λ(t) for 0 ≤ t ≤ 1 We consider the following domain, { } ∈ R2 ˆ ≥ (t, s) Re V (t, s) ςR . (79)

48 For t, s ∈ R, Re Vˆ (t, s) = −Λ(−t) − 2 Λ(−s) = Λ(t) + 2 Λ(s). 1 − − Since 2Λ( 6 ) ςR = 0.127043... < 0, the values of Λ(t) and Λ(s) must be positive when (t, s) is in the domain (79). Hence, the domain (79) is included in the following area, { } (t, s) ∈ R2 0 ≤ t ≤ 0.5, 0 ≤ s ≤ 0.5 .

We assume that 0 ≤ t ≤ 0.5 and 0 ≤ s ≤ 0.5 in the following of this proof. We consider the minimal value tmin of t such that (t, s) is in the domain (79). It satisfies 1 1 − − that Λ(tmin) + 2Λ( 6 ) = ςR . Since Λ(0.04) + 2Λ( 6 ) ςR = 0.031768... < 0, we have that 0.04 < tmin. Hence, the domain (11) is included in the area 0.04 ≤ t for some sufficiently small ε > 0. Further, in similar ways, we can show that the domain (11) is included in the area { } (t, s) ∈ R2 0.04 ≤ t ≤ 0.4, 0.05 ≤ s ≤ 0.4 for some sufficiently small ε > 0. We assume that (t, s) is in the above area in the following of this proof. By expanding Λ(t) at t = 0.3, we can show that

Λ(t) ≤ Λ(0.3) + Λ′(0.3)(t − 0.3).

If (t, s) is in the domain (79), ( ) ≤ ≤ ′ − ′ − ςR Λ(t) + 2Λ(s) Λ(0.3) + Λ (0.3)(t 0.3) + 2 Λ(0.3) + Λ (0.3)(s 0.3) . Hence, ς − 3Λ(0.3) t + 2s ≤ R + 0.9. Λ′(0.3) Further, since t ≤ 0.4, ( ) 1 ς − 3Λ(0.3) t + s ≤ R + 1.3 = 0.571668... < 0.6. 2 Λ′(0.3) Therefore, the domain (11) is included in ∆′ for some suﬃciently small ε > 0, as required.

References

[1] Andersen, J.E., Hansen, S.K., Asymptotics of the quantum invariants for surgeries on the ﬁgure 8 knot, J. Knot Theory Ramiﬁcations 15 (2006) 479–548.

[2] Dimofte T.D., Garoufalidis, S., The quantum content of the gluing equations, Geom. Topol. 17 (2013) 1253–1315.

[3] Dimofte, T., Gukov, S., Lenells, J., Zagier, D., Exact results for perturbative Chern-Simons theory with complex gauge group, Commun. Number Theory Phys. 3 (2009) 363–443.

[4] Dubois, J., Huynh, V., Yamaguchi, Y., Non-abelian Reidemeister torsion for twist knots, J. Knot Theory Ramiﬁcations 18 (2009) 303–341.

49 [5] Dubois, J., Kashaev, R., On the asymptotic expansion of the colored Jones polynomial for torus knots, Math. Ann. 339 (2007) 757–782. [6] Faddeev, L.D., Discrete Heisenberg-Weyl group and modular group, Lett. Math. Phys. 34 (1995) 249–254. [7] Faddeev, L.D., Kashaev, R.M., Strongly coupled quantum discrete Liouville theory. II. Geometric interpretation of the evolution operator, J. Phys. A 35 (2002) 4043–4048. [8] Faddeev, L.D., Kashaev, R.M., Volkov, A. Yu., Strongly coupled quantum discrete Liouville theory. I. Algebraic approach and duality, Comm. Math. Phys. 219 (2001) 199–219. [9] Garoufalidis, S., Le, T.T.Q, On the volume conjecture for small angles, arXiv:math/0502163.

[10] Garoufalidis, S., Thurston, D.P., Zickert, C.K., The complex volume of SL(n, C)-representations of 3-manifolds, Duke Math. J. 164 (2015) 2099–2160.

[11] Gukov, S., Three-dimensional quantum gravity, Chern-Simons theory, and the A-polynomial, Comm. Math. Phys. 255 (2005) 577–627.

[12] Gukov, S., Murakami, H., SL(2, C) Chern-Simons theory and the asymptotic behavior of the colored Jones polynomial, Lett. Math. Phys. 86 (2008) 79–98.

[13] Hansen, S.K., Takata, T., Quantum invariants of Seifert 3-manifolds and their asymptotic expan- sions, Invariants of knots and 3-manifolds (Kyoto, 2001), 69–87, Geom. Topol. Monogr. 4, Geom. Topol. Publ., Coventry, 2002.

[14] , Reshetikhin-Turaev invariants of Seifert 3-manifolds for classical simple Lie algebras, J. Knot Theory Ramiﬁcations 13 (2004) 617–668.

[15] Hikami, K., Quantum invariant for torus link and modular forms, Comm. Math. Phys. 246 (2004) 403–426.

[16] Kashaev, R.M., Quantum dilogarithm as a 6j-symbol, Modern Phys. Lett. A9 (1994) 3757–3768. [17] , A link invariant from quantum dilogarithm, Mod. Phys. Lett. A10 (1995) 1409–1418.

[18] , The hyperbolic volume of knots from the quantum dilogarithm, Lett. Math. Phys. 39 (1997) 269–275.

[19] Kashaev, R.M., Tirkkonen, O., A proof of the volume conjecture on torus knots (Russian), Zap. Nauchn. Sem. S.-Peterburg. Otdel. Mat. Inst. Steklov. (POMI) 269 (2000), Vopr. Kvant. Teor. Polya i Stat. Fiz. 16, 262–268, 370; translation in J. Math. Sci. (N.Y.) 115 (2003) 2033–2036.

[20] Kashaev, R.M., Yokota, Y., On the volume conjecture for the knot 52, preprint. (The authors’ affiliations: “Section de Mathématiques,Universitéde Genève, 2-4, rue de Lièvre,1211 Genève 4, Suisse” and “Department of Mathematics and Information Sciences, Tokyo Metropolitan University, Tokyo, 192-0397, Japan”.) [21] Kirk, P., Klassen, E., Chern-Simons invariants of 3-manifolds decomposed along tori and the circle bundle over the representation space of T 2, Comm. Math. Phys. 153 (1993) 521–557. [22] Murakami, H., An introduction to the volume conjecture, Interactions between hyperbolic geometry, quantum topology and number theory, 1–40, Contemp. Math. 541, Amer. Math. Soc., Providence, RI, 2011.

[23] , The coloured Jones polynomial, the Chern-Simons invariant, and the Reidemeister torsion of the ﬁgure-eight knot, J. Topol. 6 (2013) 193–216.

50 [24] Murakami, H., Murakami, J., The colored Jones polynomials and the simplicial volume of a knot, Acta Math. 186 (2001) 85–104. [25] Murakami, H., Murakami, J., Okamoto, M., Takata, T., Yokota, Y., Kashaev’s conjecture and the Chern-Simons invariants of knots and links, Experiment. Math. 11 (2002) 427–435. [26] Ohtsuki, T., On the asymptotic expansion of the Kashaev invariant of the knots with 7 crossings, in preparation. [27] Ohtsuki, T., Yokota, Y., On the asymptotic expansion of the Kashaev invariant of the knots with 6 crossings, preprint, 2011. (The second author’s aﬃliation: Department of Mathematics and Infor- mation Sciences, Tokyo Metropolitan University, Tokyo, 192-0397, Japan.)

[28] Ohtsuki, T., Takata, T., On the Kashaev invariant and the twisted Reidemeister torsion of two-bridge knots, Geom. Topol. 19 (2015) 853–952.

[29] Porti, J., Torsion de Reidemeister pour les vari´et´eshyperboliques, Mem. Amer. Math. Soc. 128 (1997), no. 612, x+139 pp.

[30] Rozansky, L., A large k asymptotics of Witten’s invariant of Seifert manifolds, Proceedings of the Conference on Quantum Topology (Manhattan, KS, 1993), 307–354, World Sci. Publ., River Edge, NJ, 1994. [31] , A large k asymptotics of Witten’s invariant of Seifert manifolds, Comm. Math. Phys. 171 (1995) 279–322.

[32] , Witten’s invariant of 3-dimensional manifolds: loop expansion and surgery calculus, Knots and applications, 271–299, Ser. Knots Everything 6, World Sci. Publ., River Edge, NJ, 1995.

[33] Stein, E.M., Weiss, G., Introduction to Fourier analysis on Euclidean spaces, Princeton Mathematical Series 32. Princeton University Press, Princeton, N.J., 1971.

[34] Takata, T., in preparation. (The author’s aﬃliation: Faculty of Mathematics, Kyushu University, Fukuoka, 819-0395, Japan.)

[35] Thurston, D.P., Hyperbolic volume and the Jones polynomial, Notes accompanying lectures at the summer school on quantum invariants of knots and three-manifolds, Joseph Fourier Institute, Uni- versity of Grenoble, org. C. Lescop, June, 1999. [36] van der Veen, R., Proof of the volume conjecture for Whitehead chains, Acta Math. Vietnam. 33 (2008) 421–431. [37] , A cabling formula for the colored Jones polynomial, arXiv:0807.2679.

[38] Witten, E., Analytic continuation of Chern-Simons theory, Chern-Simons gauge theory: 20 years after, 347–446, AMS/IP Stud. Adv. Math. 50, Amer. Math. Soc., Providence, RI, 2011.

[39] Wong, R., Asymptotic approximations of integrals, Computer Science and Scientiﬁc Computing. Academic Press, Inc., Boston, MA, 1989.

[40] Woronowicz, S.L., Quantum exponential function, Rev. Math. Phys. 12 (2000) 873–920. [41] Yamazaki, M., Yokota, Y., On the limit of the colored Jones polynomial of a non-simple link, Tokyo J. Math. 33 (2010) 537–551. [42] Yokota, Y., On the volume conjecture for hyperbolic knots, math.QA/0009165.

51 [43] , From the Jones polynomial to the A-polynomial of hyperbolic knots, Proceedings of the Winter Workshop of Topology/Workshop of Topology and Computer (Sendai, 2002/Nara, 2001). Interdiscip. Inform. Sci. 9 (2003) 11–21.

[44] , On the Kashaev invariant of twist knots, “Intelligence of Low-dimensional Topology” (edited by T. Ohtsuki and M. Wakui), RIMS K¯oky¯uroku 1766 (2011) 45–51.

[45] , On the complex volume of hyperbolic knots, J. Knot Theory Ramiﬁcations 20 (2011) 955– 976.

[46] Zagier, D., Quantum modular forms, Quanta of maths, 659–675, Clay Math. Proc. 11, Amer. Math. Soc., Providence, RI, 2010.

[47] Zheng, H., Proof of the volume conjecture for Whitehead doubles of a family of torus knots, Chin. Ann. Math. Ser. B 28 (2007) 375–388.

Research Institute for Mathematical Sciences, Kyoto University, Sakyo-ku, Kyoto, 606-8502, Japan E-mail address: [email protected]